Three Dimensional Geometry

Given, $G=(1,0,1), A(1,-4,2)$ and

$ B=(3,1,0), $

then $C=(-1,3,1)$

$ \begin{gathered} A G^2=(17) \text { and } C G^2=13 \\ A G^2+C G^2=17+13=30 \\ B G^2=6 \\ \because A G^2+C G^2=5 \cdot B G^2 \end{gathered} $

We have, $(3,4, \infty)$

Distance of point to $X$-axis $=\sqrt{16+\alpha^2}$

Distance of point to $Y$-axis $=\sqrt{9+\alpha^2}$

Distance of point to $Z$-axis $=\sqrt{9+16}=5$

Sum $=S=\sqrt{16+\alpha^2}+\sqrt{9+\alpha^2}+5$

The sum $S$ is minimum when $\alpha=0$ because $\sqrt{16+\alpha^2}$ and $\sqrt{9+\alpha^2}$ are minimum when $\alpha=0$

$\because \quad \sec 0=1$

Let equation of required plane is

$ a(x-2)+b(y+1)+c(z-3)=0 $

Now, the plane is perpendicular to

$ 3 x-2 y+z=8 \text { and } x+y+z=6 $

$ \left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & -2 & 1 \\ 1 & 1 & 1 \end{array}\right|=\hat{\mathbf{i}}(-3)-\hat{\mathbf{j}}(2)+k(5) $

$\because a, b$ and $c$ are $-3,-2$ and 5

$ \begin{aligned} & -3(x-2)-2(y+1)+5(z-3)=0 \\ \Rightarrow & -3 x-2 y+5 z+6-2-15=0 \\ \Rightarrow & -3 x-2 y+5 z-11=0 \\ \Rightarrow & -3 x-2 y+5 z=11 \\ \Rightarrow & -\frac{3}{11} x-\frac{2}{11} y+\frac{5}{11} z=1 \\ & l=-\frac{3}{11}, m=-\frac{2}{11}, n=\frac{5}{11} \\ \therefore & 4 m+2 n-3 l=-\frac{8}{11}+\frac{10}{11}+\frac{9}{11}=\frac{11}{11}=1 \end{aligned} $

To find the equation of the plane $ \pi $ that passes through the points $ (2, -3, 0) $ and $ (3, 0, 4) $, and is parallel to the vector $ \langle 2, 3, -4 \rangle $, we start with the general equation of a plane:

$ a(x - 2) + b(y + 3) + c(z - 0) = 0. $

Substituting the point $ (3, 0, 4) $ into the plane equation gives:

$ a(1) + b(3) + c(4) = 0 \quad \text{(i)}.$

Since the plane is parallel to the vector $ \langle 2, 3, -4 \rangle $, the plane's normal vector must be orthogonal to $ \langle 2, 3, -4 \rangle $. Thus,

$ 2a + 3b - 4c = 0 \quad \text{(ii)}.$

To find $ a, b, c $ such that both conditions are satisfied, compare ratios:

$ \frac{a}{-12} = \frac{-b}{-4} = \frac{c}{3}. $

Simplifying gives:

$ \frac{a}{-24} = \frac{b}{12} = \frac{c}{-3}. $

From this, solve for proportional constants:

$ \frac{a}{8} = \frac{b}{-4} = \frac{c}{1}. $

Choosing $ c = 1 $, we find $ a = 8 $ and $ b = -4 $.

The equation of the plane becomes:

$ 8(x - 2) - 4(y + 3) + z = 0, $

or simplified:

$ 8x - 4y + z - 28 = 0. $

Next, find the line equation that passes through points $ (1, 2, 0) $ and $ (0, 1, -2) $. The direction vector of the line is $ \langle -1, -1, -2 \rangle $.

The parametric equation of the line is:

$ \frac{x-1}{-1} = \frac{y-2}{-1} = \frac{z-0}{-2} = \lambda. $

Therefore, the line is:

$ P(x, y, z) = (-\lambda + 1, -\lambda + 2, -2\lambda). $

Substitute these into the plane equation:

$ 8(-\lambda + 1) - 4(-\lambda + 2) + (-2\lambda) - 28 = 0. $

Simplifying:

$ -8\lambda + 8 + 4\lambda - 8 - 2\lambda - 28 = 0, $

$ -6\lambda = 28 \quad \Rightarrow \quad \lambda = -\frac{14}{3}. $

Substituting $ \lambda = -\frac{14}{3} $ into the line equations gives the intersection point $ P(a, b, c) = \left(\frac{17}{3}, \frac{20}{3}, \frac{28}{3}\right)$.

Finally, $ a + b + 2c $ is:

$ a + b + 2c = \frac{17}{3} + \frac{20}{3} + \frac{56}{3} = \frac{93}{3} = 31. $

Equation of plane passing through the point of intersection of $ x-y+z-5=0 $ and $ 2 x+y-z-3=0 $ is $ x-y+z-5+\lambda(2 x+y-z-3)=0 $ $ \pi $ is passing through $ (0,1,2) $.

$ \begin{array}{l} 0-1+2-5+\lambda(0+1-2-3)=0 \\\\ \Rightarrow \quad-4+\lambda(-4)=0 \\\\ \Rightarrow \quad-4-4 \lambda=0 \\\\ \Rightarrow \quad 1+\lambda=0 \\\\ \Rightarrow \quad \lambda=-1 \\\\ \Rightarrow x-y+z-5-1(2 x+y-z-3)=0 \\\\ -x-2 y+2 z-2=0 \\\\ x+2 y-2 z+2=0 \\\\ r \cdot(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})=m \\\\ a=1, b=2, c=-2 \\\\ \because \quad \frac{b c}{a^{2}}=-\frac{4}{1} \\\\ \therefore \quad \frac{b c}{a^{2}}=-4 \end{array} $

$ \mathrm{AB}=\mathrm{OB}-\mathrm{OA} =3 \hat{\mathbf{i}}+(b-4) \hat{\mathbf{j}}+(3-a) \hat{\mathbf{k}} $

$ \mathbf{C D}=\mathbf{O D}-\mathbf{O C}=(-5-C) \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} $

$ \mathbf{A B} $ and $ \mathbf{C D} $ are collinear

$ \frac{3}{-5-C}=\frac{b-4}{6}=\frac{3-a}{-3}=\lambda $

$ -5-C=\frac{3}{\lambda}, b-4=6 \lambda $ and $ 3-a=-3 \lambda $

$ c=-5-\frac{3}{\lambda}, b=6 \lambda+4 $ and $ a=3+3 \lambda $

$ a+b+c=-5-\frac{3}{\lambda}+6 \lambda+4+3+3 \lambda $

$ a+b+c=2+9 \lambda-\frac{3}{\lambda} $

By intersection if we put $ \lambda=1 $ then, $ a+b+c=2+9-3=8 $

Direction ratio of

$ \begin{aligned} \mathrm{AB} \equiv(2-1,-1+2,2-1) =(1,1,1) \end{aligned} $

Equation of $ A B $

$ \begin{array}{l} \frac{x-1}{1}=\frac{y+2}{1}=\frac{z-1}{1}=D(\text { say }) \\\\ D \equiv(\lambda+1, \lambda-2, \lambda+1)=D(\alpha, \beta, \gamma) \\\\ \alpha=\lambda+1, \beta=\lambda-2, \gamma=\lambda+1 \end{array} $

Direction ratio of $ C D=\alpha-1, \beta-2, \gamma-3 $ $ A B \perp C D $

$ \alpha-1+\beta-2+\gamma-3=0 $

$ \alpha+\beta+\gamma=6 $

$ 3 \lambda=6 \Rightarrow \lambda=2 $

$ \alpha=3, \beta=0, \gamma=3 $

$ \alpha^{2}+\beta^{2}+\gamma^{2}=9+0+9=18 $

Direction ratio of $ P M=3,1,-5 $

Let equation of plane passing through

$ \begin{array}{l} \left(x_{1}, y_{1}, z_{1}\right) \\\\ 3\left(x-x_{1}\right)+1\left(y-y_{1}\right)-5\left(z-z_{1}\right)=0 \\\\ 3 x+y-5 z=3 x_{1}+y_{1}-5 z_{1} \end{array} $

$ M(1,0,-2) $ lies on the plane

$ \begin{aligned} & 13=3 x_{1}+y_{1}-5 z_{1} \\\\ \Rightarrow \quad & 3 x+y-5 z=13 \end{aligned} $

intercept on axis is

$ \begin{gathered} \frac{13}{3}, 13, \frac{-13}{5} \\\\ 3 a+b+5 c=13+13-13=13 \end{gathered} $

To find the unit vector normal to the plane π, we first calculate the cross product of the vectors $\hat{\mathbf{i}}+3 \hat{\mathbf{k}}$ and $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$. The cross product is calculated as follows:

$ \mathbf{n} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 0 & 3 \\ 2 & 1 & -1 \end{vmatrix} = (-3)\hat{\mathbf{i}} + 7\hat{\mathbf{j}} + \hat{\mathbf{k}} $

Now, we find the unit vector $\mathbf{n}_{\text{unit}}$ by dividing this normal vector by its magnitude. The magnitude of $-3\hat{\mathbf{i}} + 7\hat{\mathbf{j}} + \hat{\mathbf{k}}$ is:

$ \sqrt{(-3)^2 + 7^2 + 1^2} = \sqrt{9 + 49 + 1} = \sqrt{59} $

Thus, the unit vector is:

$ \mathbf{n}_{\text{unit}} = \frac{1}{\sqrt{59}}(-3 \hat{\mathbf{i}} + 7 \hat{\mathbf{j}} + \hat{\mathbf{k}}) $

The equation of the plane is given by:

$ -3x + 7y + z + d = 0 $

Since the plane passes through the point $(-3, 7, 1)$, we substitute these coordinates into the plane’s equation:

$ -3(-3) + 7(7) + 1 + d = 0 \\ 9 + 49 + 1 + d = 0 \\ d = -59 $

Therefore, the equation of the plane is:

$ -3x + 7y + z - 59 = 0 $

The perpendicular distance $p$ from the origin to the plane can be calculated using the formula for the distance from a point to a plane:

$ p = \frac{|0(-3) + 0(7) + 0(1) - 59|}{\sqrt{(-3)^2 + 7^2 + 1^2}} = \frac{59}{\sqrt{59}} = \sqrt{59} $

Given $p$, we calculate:

$ \sqrt{p^2 + 5} = \sqrt{(\sqrt{59})^2 + 5} = \sqrt{59 + 5} = \sqrt{64} = 8 $

To find the harmonic conjugate $ Q(\alpha, \beta, \gamma) $ of the point $ P(2,3,4) $ with respect to the line segment joining points $ A(3,-2,2) $ and $ B(6,-17,-4) $, we start by noting that $ P $ divides $ AB $ in some ratio $ \lambda : 1 $.

First, let's determine this ratio using the x-coordinate:

$ 2 = \frac{6\lambda + 3}{\lambda + 1} $

Solving for $ \lambda $:

$ 2(\lambda + 1) = 6\lambda + 3 \\ 2\lambda + 2 = 6\lambda + 3 \\ 2\lambda - 6\lambda = 3 - 2 \\ -4\lambda = 1 \\ \lambda = -\frac{1}{4} $

This indicates that $ P $ divides $ AB $ in the ratio $-\frac{1}{4} : 1$.

For the harmonic conjugate $ Q $, the ratio changes sign, yielding a division ratio of $ 1 : 4 $ (a harmonized ratio for $-\frac{1}{4} : 1$).

We calculate the coordinates of $ Q $ using the section formula in the new ratio $ 1:4 $:

$ Q = \left( \frac{1 \cdot 6 + 4 \cdot 3}{1+4}, \frac{1 \cdot (-17) + 4 \cdot (-2)}{1+4}, \frac{1 \cdot (-4) + 4 \cdot 2}{1+4} \right) $

Calculating each coordinate:

$ x = \frac{6 + 12}{5} = \frac{18}{5} $

$ y = \frac{-17 - 8}{5} = \frac{-25}{5} = -5 $

$ z = \frac{-4 + 8}{5} = \frac{4}{5} $

Thus, $ Q = \left( \frac{18}{5}, -5, \frac{4}{5} \right) $.

Finally, we find $\alpha + \beta + \gamma$:

$ \alpha + \beta + \gamma = \frac{18}{5} - 5 + \frac{4}{5} = \frac{18}{5} - \frac{25}{5} + \frac{4}{5} = -\frac{3}{5} $

The line $ L $ is the intersection of the two planes given by the equations $ x + 2y + 2z = 15 $ and $ x - y + z = 4 $. The direction ratios of the line $ L $ are determined by the cross product of the normal vectors of these planes.

The normal vector of the first plane is $ \langle 1, 2, 2 \rangle $ and for the second plane is $ \langle 1, -1, 1 \rangle $. We calculate the cross product of these two vectors, represented by the determinant:

$ \left| \begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 2 \\ 1 & -1 & 1 \end{array} \right| $

This determinant is computed as follows:

$ 4\hat{\mathbf{i}} + \hat{\mathbf{j}} - 3\hat{\mathbf{k}} $

Thus, the direction ratios $ (a, b, c) $ of line $ L $ are $ (4, 1, -3) $.

Finally, we need to find the value of:

$ \frac{a^2 + b^2 + c^2}{b^2} $

Substituting the values, we get:

$ \frac{4^2 + 1^2 + (-3)^2}{1^2} = \frac{16 + 1 + 9}{1} = 26 $

Therefore, the result is 26.

Given that the foot of the perpendicular from $A(1,2,2)$ to the plane $x + 2y + 2z - 5 = 0$ is $B(\alpha, \beta, \gamma)$, we can begin by determining the direction ratios of the perpendicular.

Start by using the formula for the foot of the perpendicular:

$ \frac{\alpha - 1}{1} = \frac{\beta - 2}{2} = \frac{\gamma - 2}{2} = \frac{-(1 + 4 + 4 - 5)}{1 + 4 + 4} $

Simplifying, we get:

$ \Rightarrow \frac{\alpha - 1}{1} = \frac{\beta - 2}{2} = \frac{\gamma - 2}{2} = \frac{-4}{9} $

Solving these equations, we find:

$ \alpha = \frac{5}{9},\ \ \beta = \frac{-8}{9} + 2 = \frac{10}{9},\ \ \gamma = \frac{-8}{9} + 2 = \frac{10}{9} $

Now, calculate $\pi(A)$:

$ \pi(A) = 1 + 2 \times 2 + 2 \times 2 + 5 = 14 $

Next, find $\pi(B)$:

$ \pi(B) = \frac{5}{9} + 2 \left(\frac{10}{9}\right) + 2 \left(\frac{10}{9}\right) + 5 = \frac{5}{9} + \frac{20}{9} + \frac{20}{9} + 5 = 10 $

Finally, compute the ratio $\frac{-\pi(A)}{\pi(B)}$:

$ \frac{-\pi(A)}{\pi(B)} = \frac{-14}{10} = \frac{-7}{5} $

Equation of plane $ \pi_{1} $ is

$ \begin{array}{r} 1(x-3)+2(y+7)-2(z-5)=0 \\\\ x+2 y-2 z-3+14+10=0 \\\\ x+2 y-2 z+21=0 \quad \ldots \text { (i) } \end{array} $

Equation of plane $ \pi_{2} $ is

$ \begin{array}{c} 3(x-2)+2(y-7)+6(z+8)=0 \\\\ 3 x+2 y+6 z-6-14+48=0 \\\\ 3 x+2 y+6 z+28=0\quad \ldots \text { (ii) } \\\\ P_{1}=\frac{21}{\sqrt{1+4+4}}=7 \text { units } \\\\ P_{2}=\frac{28}{\sqrt{9+4+36}}=4 \text { units } \end{array} $

$ P_{1}-P_{2}=(7-4) $ units $ =3 $ units

Given, $ A \equiv(2,3, K), B \equiv(-1, K,-1) $

and $ C \equiv(4,-3,2) $

$ \begin{array}{l} A B=A C \Rightarrow|A B|^{2}=|A C|^{2} \\\\ 9+(K-3)^{2}+(1+K)^{2}=4+36+(K-2)^{2} \\\\ 9+K^{2}+9-6 K+1+K^{2}+2 K \\\\ \qquad 40+K^{2}+4-4 K \\\\ 2 K^{2}-4 K+19-40-K^{2}+4 K-4=0 \\\\ K^{2}=44-19=25 \Rightarrow K=5 \\\\ A \equiv(2,3,5), B \equiv(-1,5,-1), C \equiv(4,-3,2) \\\\ A C=A B=\sqrt{9+4+36}=7 \\\\ B C=\sqrt{25+64+9}=7 \sqrt{2} \end{array} $

Side are $ (7,7,7 \sqrt{2}) $

Equation of plane

$ \begin{array}{l} \left|\begin{array}{ccc} x-1 & y & z+2 \\\\ 3-1 & -1 & 2+2 \\\\ -1 & -3 & 6 \end{array}\right|=0 \\\\ \left|\begin{array}{ccc} x-1 & y & z+2 \\\\ 2 & -1 & 4 \\\\ -1 & -3 & 6 \end{array}\right|=0 \\\\ \Rightarrow(x-1)(-6+12)-y(12+4) \\\\ +(z+2)(-6-1)=0 \\\\ \begin{array}{r} \Rightarrow 6 x-6-16 y-7 z-14=0 \\\\ 6 x-16 y-7 z=20 \end{array} \\\\ \begin{array}{r} x \text {-intercept }=\frac{20}{6}=a \end{array} \\\\ \begin{array}{r} y \text {-intercept }=\frac{-20}{16}=b \end{array} \\\\ z \text {-intercept }=\frac{-20}{7}=c \\\\ 3 a+4 b+7 c=10-5-20=-15 \end{array} $

We have,

$ \begin{aligned} & \mathrm{OA}=\hat{\mathbf{i}}+\hat{\mathbf{j}}, \mathrm{OB}=\hat{\mathbf{j}}+\hat{\mathbf{k}} \\\\ & \mathrm{OC}=\hat{\mathbf{k}}+\hat{\mathbf{i}}, \mathrm{OD}=\hat{\mathbf{i}}-\hat{\mathbf{j}} \text { and } \mathrm{OE}=\hat{\mathbf{j}}-\hat{\mathbf{k}} \\\\ & \therefore \mathrm{CD}=(\hat{\mathbf{i}}-\hat{\mathbf{j}})-(\hat{\mathbf{k}}+\hat{\mathbf{i}})=-\hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} & \mathbf{C E}=(\hat{\mathbf{j}}-\hat{\mathbf{k}})-(\hat{\mathbf{k}}+\hat{\mathbf{i}})=-\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & \mathbf{D E}=(\hat{\mathbf{j}}-\hat{\mathbf{k}})-(\hat{\mathbf{i}}-\hat{\mathbf{j}})=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & \mathbf{n}=\mathbf{C D} \times \mathbf{C E}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & -1 & -1 \\ -1 & 1 & -2 \end{array}\right|=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned} $

Vector equation of palne,

$ \begin{aligned} & \mathbf{r} \cdot(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=(\hat{\mathbf{k}}+\hat{\mathbf{i}}) \cdot(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ \Rightarrow & \mathbf{r} \cdot(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=2 \quad \ldots \text { (i) } \end{aligned} $

Vector equation line,

$ \begin{aligned} \mathbf{r} & =(\hat{\mathbf{i}}+\hat{\mathbf{j}})+t[(\hat{\mathbf{j}}+\hat{\mathbf{k}})-(\hat{\mathbf{i}}+\hat{\mathbf{j}})] \\ \Rightarrow \mathbf{r} & =(\mathrm{l}-t \hat{\mathbf{i}}+\hat{\mathbf{j}}+t \hat{\mathbf{k}} \quad \ldots \text { (ii) } \end{aligned} $

From Eqs. (i) and (ii), we get

$ \begin{aligned} & {[(1-t) \hat{\mathbf{i}}+\hat{\mathbf{j}}+t \hat{\mathbf{k}}] \cdot(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=2 } \\ \Rightarrow & 3(1-t)+1-t=2 \Rightarrow 3-3 t+1-t=2 \\ \Rightarrow & \quad-4 t=-2 \Rightarrow t=\frac{1}{2} \end{aligned} $

On putting $t=\frac{1}{2}$ in Eq. (ii), we get

$ \begin{aligned} \mathbf{r} & =\left(1-\frac{1}{2}\right) \hat{\mathbf{i}}+\hat{\mathbf{j}}+\frac{1}{2} \hat{\mathbf{k}} \\ & =\frac{1}{2} \hat{\mathbf{i}}+\hat{\mathbf{j}}+\frac{1}{2} \hat{\mathbf{k}}, \text { which is the required } \end{aligned} $

point of intersection.

Normal vector to the plane

$ x+y-z+4=0 \text { is } \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} $

Normal vector to the plane

$ 2 x-y+z+1=0 \text { is } 2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} $

Given planes are perpendicular to the plane ( $\pi$ )

$\therefore$ Normal vector to the plane $(\pi)$ is given by

$ \begin{aligned} \left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\\\ 1 & 1 & -1 \\\\ 2 & -1 & 1 \end{array}\right| & =0 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\\\ & =-3(0 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \end{aligned} $

Direction Ratio's of $-3(0 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})$ is ( $0,1,1$ )

Plane $(\pi)$ is passing through the point ( $1,2,-3$ )

$\therefore$ Plane $(\pi) 0(x-1)+1(y-2)+1(z+3)=0$

$ \begin{array}{lr} \Rightarrow & y-2+z+3=0 \\\\ \Rightarrow & 0 x+y+z+1=0 \end{array} $

Given Plane $(\pi) a x+b y+c z+1=0$

$ \begin{aligned} & \therefore \quad a=0, b=1 \text { and } c=1 \\\\ & \text { Now, } a^2+b^2+c^2=0^2+1^2+1^2 \\\\ & =0+1+1=2 \end{aligned} $

We have a variable point $P(x, y, z)$

Distance of $P$ from $X$-axis

$ =\sqrt{0^2+y^2+z^2}=\sqrt{y^2+z^2} $

and distance of $P$ from $Y Z$-plane $=|x|$

Now, $\frac{\sqrt{y^2+z^2}}{|x|}=\frac{2}{3} \quad \text { [given] }$

$ \Rightarrow \frac{y^2+z^2}{x^2}=\frac{4}{9} \Rightarrow 4 x^2-9 y^2-9 z^2=0 $

which is the equation of locus of $P$.

We have,

$ \begin{aligned} & l-m+n=0 \Rightarrow m=n+l\quad \ldots \text { (i) } \\\\ & \text { and } 2 l m-3 m n+n l=0 \\\\ & \Rightarrow 2 l(n+l)-3(n+l) n+n l=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow 2 l^2-3 n^2=0 \quad \text { [from Eq. (i)] } \\\\ & (\sqrt{2} l+\sqrt{3} n)(\sqrt{2} l-\sqrt{3} n)=0 \\\\ & \Rightarrow \quad n=\frac{\sqrt{2}}{\sqrt{3}} l, \frac{-\sqrt{2}}{\sqrt{3}} l \\\\ & \text { Since, } n=\frac{\sqrt{2}}{\sqrt{3}} l \\\\ & \Rightarrow \quad m=\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}}\right)l \quad \text { [from Eq. (i)] } \\\\ & \text { and } n=-\frac{\sqrt{2}}{\sqrt{3}} l \Rightarrow m=\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}\right) l \quad \text { [from Eq. (i)] } \end{aligned} $

So, $\frac{l}{1}=\frac{n}{\frac{\sqrt{2}}{\sqrt{3}}}=\frac{m}{\left(\frac{\sqrt{2}+\sqrt{3}}{\sqrt{3}}\right)}$ and

$ \frac{l}{1}=\frac{n}{-\frac{\sqrt{2}}{\sqrt{3}}}=\frac{m}{\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}\right)} $

$ \begin{aligned} & \text { Now, } \\\\ & \begin{aligned} & \cos \theta=\frac{11+\frac{\sqrt{2}}{\sqrt{3}} \cdot\left(-\frac{\sqrt{2}}{\sqrt{3}}\right)+\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}}\right)\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}\right)}{\sqrt{(1)^2+\left(\frac{\sqrt{2}}{\sqrt{3}}\right)^2+\left(\frac{\sqrt{2}+\sqrt{3})^2}{\sqrt{3}}\right)^2}} \\\\ &=\frac{1-\frac{2}{3}+\frac{1}{3}}{\left.\sqrt{\frac{5}{3}+\frac{5+2 \sqrt{6}}{3}} \cdot \sqrt{\frac{5}{3}+\frac{5-2 \sqrt{6}}{3}}\right)^2+\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}\right)^2} \\&=\frac{2 / 3}{\sqrt{\frac{2}{3}(5+\sqrt{6})} \cdot \sqrt{\frac{2}{3}(5-\sqrt{6})}} \end{aligned} \\\\ & \Rightarrow \cos \theta=\frac{1}{\sqrt{19}} \end{aligned} $

We have,

$ \begin{aligned} & \text { (c) We have, } \\\\ & A \equiv(5,1,2), B \equiv(3,-4,6), C \equiv(7,0,-1) \end{aligned} $

So, normal vector to plane $\pi$ is

$ \mathbf{n}=\mathbf{A B} \times \mathbf{A C} $

Here, $\mathbf{A B}=-2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$

$ A C=2 \hat{\mathbf{i}}-1 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} $

So, $\mathbf{n}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\\\ -2 & -5 & 4 \\\\ 2 & -1 & -3\end{array}\right|$

$ \mathbf{n}=19 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+12 \hat{\mathbf{k}} $

Now, equations of plane through point $A$ with normal vector $(19,2,12)$ is

$ 19(x-5)+2(y-1)+12(z-2)=0 $

So, $P=\frac{|-95-2-24|}{\sqrt{361+4+144}}=\frac{121}{\sqrt{509}}$

Direction cosines of $\mathbf{n}$ is $l=\frac{19}{\sqrt{509}}$,

$ m=\frac{2}{\sqrt{509}}, n=\frac{12}{\sqrt{509}} $

Hence,

$ |3 l+2 m+5 n|=\left|\frac{57+4+60}{\sqrt{509}}\right|=\frac{121}{\sqrt{509}}=P $

Given the points $A(4,7,8)$, $B(2,3,4)$, and $C(2,5,7)$, we need to find the length of the internal bisector of angle $A$ in $\triangle ABC$.

Let's denote the intersection of the angle bisector at $A$ with the line segment $BC$ as point $D$. The angle bisector divides $BC$ in the ratio $|AB|:|AC|$.

First, calculate the vectors $\mathbf{AB}$ and $\mathbf{AC}$:

$ \begin{aligned} \mathbf{AB} &= B - A = (2 - 4)\mathbf{i} + (3 - 7)\mathbf{j} + (4 - 8)\mathbf{k} = -2\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}, \\ \mathbf{AC} &= C - A = (2 - 4)\mathbf{i} + (5 - 7)\mathbf{j} + (7 - 8)\mathbf{k} = -2\mathbf{i} - 2\mathbf{j} - \mathbf{k}. \end{aligned} $

Now, calculate the magnitudes $|\mathbf{AB}|$ and $|\mathbf{AC}|$:

$ \begin{aligned} |\mathbf{AB}| &= \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = 6, \\ |\mathbf{AC}| &= \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = 3. \end{aligned} $

Using these magnitudes, we find the position vector of $D$:

$ \begin{aligned} \text{Position vector of } D &= \frac{|\mathbf{AC}| \cdot \text{PV of } B + |\mathbf{AB}| \cdot \text{PV of } C}{|\mathbf{AB}| + |\mathbf{AC}|} \\ &= \frac{3(2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) + 6(2\mathbf{i} + 5\mathbf{j} + 7\mathbf{k})}{9} \\ &= \frac{6\mathbf{i} + 9\mathbf{j} + 12\mathbf{k} + 12\mathbf{i} + 30\mathbf{j} + 42\mathbf{k}}{9} \\ &= \frac{18\mathbf{i} + 39\mathbf{j} + 54\mathbf{k}}{9} \\ &= 2\mathbf{i} + \frac{13}{3}\mathbf{j} + 6\mathbf{k}. \end{aligned} $

The vector $\mathbf{AD}$ is:

$ \mathbf{AD} = D - A = (2 - 4)\mathbf{i} + \left(\frac{13}{3} - 7\right)\mathbf{j} + (6 - 8)\mathbf{k} = -2\mathbf{i} - \frac{8}{3}\mathbf{j} - 2\mathbf{k}. $

Calculate the length $|\mathbf{AD}|$:

$ \begin{aligned} |\mathbf{AD}| &= \sqrt{(-2)^2 + \left(-\frac{8}{3}\right)^2 + (-2)^2} \\ &= \sqrt{4 + \frac{64}{9} + 4} \\ &= \sqrt{\frac{36}{9} + \frac{64}{9} + \frac{36}{9}} \\ &= \sqrt{\frac{136}{9}} \\ &= \frac{\sqrt{136}}{3} \\ &= \frac{2\sqrt{34}}{3} \text{ units}. \end{aligned} $

We have, $l+m+n=0$

and $m n-2 l m-2 n l=0$

$ \begin{aligned} & \Rightarrow \quad m n-21(m+n)=0 \\ & \Rightarrow \quad m n+2(m+n)(m+n)=0 \\ & \Rightarrow \quad 2\left(m^2+n^2+2 m n\right)+m n=0 \\ & \Rightarrow \quad 2 m^2+2 n^2+5 m n=0 \\ & \Rightarrow \quad 2 m^2+5 m n+2 n^2=0 \\ & \Rightarrow \quad 2 m^2+4 m n+m n+2 n^2=0 \\ & \Rightarrow \quad 2 m(m+2 n)+n(m+2 n)=0 \\ & \Rightarrow \quad(m+2 n)(2 m+n)=0 \\ & \Rightarrow \quad m+2 n=0 \text { or } 2 m+n=0 \end{aligned} $

When, $m+2 n=0 \Rightarrow m=-2 n$

Then, $l+(-2 n)+n=0 \Rightarrow l=n$

$ \Rightarrow \quad \frac{l}{1}=\frac{m}{-2}=\frac{n}{1} $

When, $2 m+n=0 \Rightarrow 2 m=-n$

Then, $l+m+(-2 m)=0$

$ \begin{aligned} & \Rightarrow \quad l-m=0 \Rightarrow l=m \\ & \Rightarrow \quad \frac{l}{1}=\frac{m}{1}=\frac{n}{-2} \\ & \therefore \cos \theta=\frac{|1-2-2|}{\sqrt{1+4+1} \sqrt{1+1+4}}=\frac{1}{2} \\ & \Rightarrow \quad \theta=\frac{\pi}{3} \end{aligned} $

$ \begin{aligned} &\text { Given, equation of lines }\\ &\begin{aligned} & \frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2} \\ & \text { and plane } 2 x-y+\sqrt{\lambda} z+4=0 \\ & \therefore \sin \theta=\frac{1 \times 2+2 \times(-1)+2 \times \sqrt{\lambda}}{\sqrt{1+4+4} \sqrt{4+1+\lambda}} \\ & \Rightarrow \quad \frac{1}{3}=\frac{2-2+2 \sqrt{\lambda}}{3 \times \sqrt{5+\lambda}} \\ & \Rightarrow \quad \frac{1}{3}=\frac{2 \sqrt{\lambda}}{3 \sqrt{5+\lambda}} \\ & \Rightarrow \quad \sqrt{5+\lambda}=2 \sqrt{\lambda} \\ & \Rightarrow \quad 5+\lambda=4 \lambda \\ & \Rightarrow \quad \lambda=\frac{5}{3} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Given, } A(1,2,3), B(3,4,7) \text {, } \\ & \begin{aligned} & C(-3,-2,-5) \\ & \Rightarrow \quad \mathbf{A B}=\mathbf{O B}-\mathbf{O A} \\ &=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\ & \mathbf{B C}=\mathbf{O C}-\mathbf{O B} \\ &=-6 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-12 \hat{\mathbf{k}} \end{aligned} \end{aligned} $

Let $A, B$ and $C$ be in a line

Then,

$ \begin{aligned} & |\mathbf{A B} \cdot \mathbf{B C}|=|\mathbf{A B}| \times|\mathbf{B C}| \\ & \text { So, }|\mathbf{A B} \cdot \mathbf{B C}|=|-12-12-48|=+72 \\ & \text { Now, }|\mathbf{A B}|=\sqrt{4+4+16}=\sqrt{24} \\ & |\mathbf{B C}|=\sqrt{36+36+144}=\sqrt{216} \\ & |\mathbf{A B}| \times|\mathbf{B C}|=\sqrt{24} \times \sqrt{216}=72 \end{aligned} $

So, $A B C$ are collinear.

Given that

$ P(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}), Q(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}), R(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) $

and $S(2 \hat{\mathbf{i}}+\hat{\mathbf{j}})$ are vertices of a tetrahedron

volume $=\frac{1}{6}\left[\begin{array}{lll}\mathbf{P Q} & \mathbf{P R} & \mathbf{P S}\end{array}\right]$

$ \begin{aligned} & =\frac{1}{6}[(2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \quad(2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \quad(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})] \\ & =\frac{1}{6}\left|\begin{array}{lll} 0 & 2 & 2 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{array}\right|=\frac{1}{6}[-2(0-3)+2(0-2)] \\ & =\frac{1}{6}(6-4)=\frac{2}{6}=\frac{1}{3} \end{aligned} $

$ \begin{aligned} &\text { Let the direction cosine of line be }\\ &\begin{aligned} &(l, m, n) \\ & \text { Then, } l=\cos \alpha \\ &\,\,\,\,\,\,\,\,\,\,\, m=\cos \beta \\ & \text { and } \quad n=\cos \gamma \end{aligned} \end{aligned} $

$ \begin{array}{ll} \text { Given, } \beta=\frac{\pi}{3}, \gamma=\frac{\pi}{4} \\ \Rightarrow m=\cos \frac{\pi}{3}=\frac{1}{2} \\ \text { and } n=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \\ \therefore l^2+m^2+n^2=1 \\ \Rightarrow l^2+\frac{1}{4}+\frac{1}{2}=1 \Rightarrow l^2=1-\frac{3}{4}=\frac{1}{4} \\ l= \pm \frac{1}{2} \\ \Rightarrow (l, m, n)=\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{\sqrt{2}}\right) \end{array} $

Direction ratio of second line is $(1,1,1)$ So, direction cosine

$ \left(l^{\prime}, m^{\prime}, n^{\prime}\right)=\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) $

Now, angle between two line

$ \begin{aligned} \cos \theta & =l^{\prime}+m m^{\prime}+n n^{\prime} \\ & =\frac{1}{2 \sqrt{3}}+\frac{1}{2 \sqrt{3}}+\frac{1}{\sqrt{6}} \end{aligned} $

$ \begin{aligned} & =\frac{1+1+\sqrt{2}}{2 \sqrt{3}}=\frac{2+\sqrt{2}}{2 \sqrt{3}}=\frac{1+\sqrt{2}}{\sqrt{6}} \\ \theta & =\cos ^{-1}\left(\frac{1+\sqrt{2}}{\sqrt{6}}\right) \end{aligned} $

Given:

A line with direction ratios $(1, 2, -1)$ has the direction cosines:

$ \left(\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}\right) $

Another line with direction ratios $(1, -2, 1)$ has the direction cosines:

$ \left(\frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right) $

A line with direction cosines $(l, m, n)$ is perpendicular to both given lines. This gives us the following conditions based on dot product orthogonality:

$ \begin{aligned} & l + 2m - n = 0, \\ & l - 2m + n = 0. \end{aligned} $

Solving these equations simultaneously, we find:

From the first equation: $ l + 2m - n = 0 $

From the second equation: $ l - 2m + n = 0 $

By adding the two equations, we get:

$ l + 2m - n + l - 2m + n = 0 \implies 2l = 0 \implies l = 0 $

Substitute $ l = 0 $ back into either equation, say $ l + 2m - n = 0 $:

$ 0 + 2m - n = 0 \implies n = 2m $

Next, using the condition for direction cosines:

$ l^2 + m^2 + n^2 = 1 $

Substitute $ l = 0 $ and $ n = 2m $:

$ 0^2 + m^2 + (2m)^2 = 1 \implies m^2 + 4m^2 = 1 \implies 5m^2 = 1 \implies m = \pm \frac{1}{\sqrt{5}} $

Also, with $ m = \pm \frac{1}{\sqrt{5}} $, it follows that:

$ n = \frac{2}{\sqrt{5}} $

Finally, calculate $(l + m + n)^2$:

$ (l + m + n)^2 = \left(0 + \frac{1}{\sqrt{5}} + \frac{2}{\sqrt{5}}\right)^2 = \left(\frac{3}{\sqrt{5}}\right)^2 = \frac{9}{5} $

Thus, $(l + m + n)^2 = \frac{9}{5}$.

$ \text { Given, } A(1,1,1) \text { and } P(-3,3,5) $

$ \begin{aligned}Mid-point of AP & =\left(\frac{1-3}{2}, \frac{1+3}{2}, \frac{1+5}{2}\right) \\ & =(-1,2,3) \end{aligned} $

and normal vector of $\overrightarrow{\mathbf{A P}}=(-4,2,4)$

The plane passing through $P$ with normal vector, then,

$ \begin{aligned} & -4\left(x-x_1\right)+2\left(y-y_1\right)+4\left(z-z_1\right)=0 \\ & \Rightarrow \quad-4 x_0+2 y+4 z=-4 x_1+2 y_1+4 z_1 \\ & \Rightarrow \quad-4 x+2 y+4 z=16+4+20=38 \\ & \Rightarrow \quad-4 x+2 y+4 z=38 \\ & \Rightarrow \quad-2 x+y+2 z=19 \end{aligned} $

Since, a plane parallel to $-2 x+y+2 z-19=0$ has same normal vector.

Thus, equation of plane passing through $(-1,2,3)$

$ \begin{aligned} & 2(x+1)-1(y-2)-2(z-3) =0 \\ \Rightarrow & 2 x-y-2 z+10 =0 \end{aligned} $

$ \begin{aligned} &\text { On comparing with the general form }\\ &\begin{gathered} a x-y+c z+d=0 \\ \therefore \quad a=2, c=-2, d=10 \\ \text { Thus, } a+c-d=2-2-10=-10 \end{gathered} \end{aligned} $

To find the distance of the point $(2,3,-5)$ from the plane given by $\hat{\mathbf{r}} \cdot (4 \hat{\mathbf{i}} - 3 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}}) = 4$, we first express the plane equation in its cartesian form:

$ 4x - 3y + 2z - 4 = 0 $

Given the formula for the distance $D$ from a point $(x_1, y_1, z_1)$ to a plane $ax + by + cz + d = 0$:

$ D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} $

We substitute the values from the point $(2, 3, -5)$ and the plane equation:

$a = 4$, $b = -3$, $c = 2$, $d = -4$

$x_1 = 2$, $y_1 = 3$, $z_1 = -5$

Plug these into the formula:

$ D = \frac{|4 \times 2 + (-3) \times 3 + 2 \times (-5) - 4|}{\sqrt{4^2 + (-3)^2 + 2^2}} $

Simplify the numerator:

$ = \frac{|8 - 9 - 10 - 4|}{\sqrt{16 + 9 + 4}} $

$ = \frac{|-15|}{\sqrt{29}} $

Which yields:

$ D = \frac{15}{\sqrt{29}} \, \text{units} $

The given points are $A(2,1,5)$, $B(3,2,3)$, and $C(4,0,4)$.

To find the orthocenter of the triangle formed by these points, we first calculate the length of each side:

Calculating $AB$:

$ \begin{aligned} AB & = \sqrt{(3-2)^2 + (2-1)^2 + (3-5)^2} \\ & = \sqrt{1^2 + 1^2 + (-2)^2} \\ & = \sqrt{1 + 1 + 4} \\ & = \sqrt{6} \end{aligned} $

Calculating $BC$:

$ \begin{aligned} BC & = \sqrt{(4-3)^2 + (0-2)^2 + (4-3)^2} \\ & = \sqrt{1^2 + (-2)^2 + 1^2} \\ & = \sqrt{1 + 4 + 1} \\ & = \sqrt{6} \end{aligned} $

Calculating $AC$:

$ \begin{aligned} AC & = \sqrt{(4-2)^2 + (0-1)^2 + (4-5)^2} \\ & = \sqrt{2^2 + (-1)^2 + (-1)^2} \\ & = \sqrt{4 + 1 + 1} \\ & = \sqrt{6} \end{aligned} $

Since all sides are equal, the triangle is equilateral. For an equilateral triangle, the orthocenter coincides with the centroid.

The centroid $P$ of the triangle is given by:

$ \begin{aligned} P &= \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right) \\ &= \left( \frac{2 + 3 + 4}{3}, \frac{1 + 2 + 0}{3}, \frac{5 + 3 + 4}{3} \right) \\ &= \left( \frac{9}{3}, \frac{3}{3}, \frac{12}{3} \right) \\ &= (3, 1, 4) \end{aligned} $

Therefore, the orthocenter of the triangle is $(3, 1, 4)$.

Given the points $ P = (0, 1, 2) $, $ Q = (4, -2, 1) $, and $ O = (0, 0, 0) $, we need to determine the angle $\angle POQ$.

First, calculate the vectors $\mathbf{OP}$ and $\mathbf{OQ}$:

$ \mathbf{OP} = 0 \hat{\mathbf{i}} + \hat{\mathbf{j}} + 2 \hat{\mathbf{k}} $

$ \mathbf{OQ} = 4 \hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + \hat{\mathbf{k}} $

The dot product of the vectors $\mathbf{OP}$ and $\mathbf{OQ}$ is given by:

$ \mathbf{OP} \cdot \mathbf{OQ} = (0 \cdot 4) + (1 \cdot -2) + (2 \cdot 1) = -2 + 2 = 0 $

The magnitudes of $\mathbf{OP}$ and $\mathbf{OQ}$ are calculated as follows:

$ |\mathbf{OP}| = \sqrt{0^2 + 1^2 + 2^2} = \sqrt{5} $

$ |\mathbf{OQ}| = \sqrt{4^2 + (-2)^2 + 1^2} = \sqrt{21} $

The angle $\theta$ between $\mathbf{OP}$ and $\mathbf{OQ}$ is found using the dot product formula:

$ \mathbf{OP} \cdot \mathbf{OQ} = |\mathbf{OP}| \, |\mathbf{OQ}| \cos \theta $

Substituting known values, we have:

$ 0 = \sqrt{5} \, \sqrt{21} \cos \theta $

Since the dot product is zero, $\cos \theta = 0$. This implies:

$ \theta = \frac{\pi}{2} $

Thus, the angle $\angle POQ$ is $\frac{\pi}{2}$.

To find the value of $ k $, we start with the equation of the plane:

$ 2x + 2y - z + k = 0 $

We know the perpendicular distance $ D $ from the point $ P(1,2,4) $ to the plane is 3. The formula for the perpendicular distance from a point $ (x_1, y_1, z_1) $ to the plane $ ax + by + cz + d = 0 $ is given by:

$ D = \left|\frac{ax_1 + by_1 + cz_1 + d}{\sqrt{a^2 + b^2 + c^2}}\right| $

Substitute the values from our plane and point into the formula. Here, $ a = 2 $, $ b = 2 $, $ c = -1 $, and $ d = k $. Also, $ x_1 = 1 $, $ y_1 = 2 $, and $ z_1 = 4 $.

$ 3 = \left|\frac{2(1) + 2(2) + (-1)(4) + k}{\sqrt{2^2 + 2^2 + (-1)^2}}\right| $

Calculate inside the absolute value:

$ 2 \times 1 + 2 \times 2 + (-1) \times 4 + k = 2 + 4 - 4 + k = 2 + k $

Calculate the denominator:

$ \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 $

The distance formula simplifies to:

$ 3 = \left|\frac{2 + k}{3}\right| $

Multiply both sides by 3 to eliminate the fraction:

$ 9 = |2 + k| $

This gives two possible equations:

$ 2 + k = 9 $

$ 2 + k = -9 $

Solve each equation for $ k $:

$ k = 9 - 2 = 7 $

$ k = -9 - 2 = -11 $

Since we're considering the positive solution that makes physical sense in most geometric contexts, we take $ k = 7 $.

We have, planes

$ \begin{aligned} & \text { } \\ & P_1=\mathbf{r} \cdot(12 \hat{i}+4 \hat{j}-3 \hat{k})=5 \text { and } \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & P_2=\mathbf{r} \cdot(5 \hat{i}+3 \hat{j}-4 \hat{k})=7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From planes (i) and (ii), we get

$ a_1=12, b_1=4, c_1=-3 \Rightarrow a_2=5, b_2=3, c_2=4 $

$ \begin{aligned} & \text { We know that } \\ & \Rightarrow \cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}} \\ & \Rightarrow \cos \theta=\frac{12 \times 5+4 \times 3-3 \times 3}{\sqrt{144+16+12} \sqrt{25+9+16}} \\ & \Rightarrow \cos \theta=\frac{60}{\sqrt{169} \sqrt{50}}=\frac{60}{13 \times 5 \sqrt{2}} \Rightarrow \cos \theta=\frac{6 \sqrt{2}}{13} \Rightarrow \theta=\cos ^{-1}\left(\frac{6 \sqrt{2}}{13}\right) \end{aligned} $

Angle between the planes is $\theta=\cos ^{-1}\left(\frac{6 \sqrt{2}}{13}\right)$

Given skew line

$ \mathbf{r}=(2 \hat{\mathbf{i}}-\hat{\mathbf{j}})+t(\hat{\mathbf{i}}+2 \hat{\mathbf{k}}) \Rightarrow \mathbf{r}=(-2 \hat{\mathbf{i}}+\hat{\mathbf{k}})+s(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}) $

We know that shortest distance between lines vector equations. $\mathbf{r}=\mathbf{a}_1+t \mathbf{b}_1 \ldots$ and $\mathbf{r}=\mathbf{a}_2+s \mathbf{b}_2 \ldots$ is

$ \left|\frac{\left(\mathbf{b}_1 \times \mathbf{b}_2\right) \cdot\left(\mathbf{a}_2-\mathbf{a}_1\right)}{\left(\mathbf{b}_1 \times \mathbf{b}_2\right)}\right| $

Now, $\mathbf{r}=(2 \hat{\mathbf{i}}-\hat{\mathbf{j}})+t(\hat{\mathbf{i}}+2 K)$ comparing with

$ \mathbf{r}=\mathbf{a}_1+t \mathbf{b}_1 \Rightarrow \mathbf{a}_1=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}, \mathbf{b}_1=\hat{\mathbf{i}}+2 \hat{\mathbf{k}} $

and $\mathbf{r}=(-2 \hat{\mathbf{i}}+\hat{\mathbf{k}})+s(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})$ comparing with

$ \begin{aligned} \mathbf{r} & =\mathbf{a}_2+t \mathbf{b}_2 \\ \therefore \quad \mathbf{a}_2 & =-2 \hat{\mathbf{i}}+\hat{\mathbf{k}}, \mathbf{b}_2=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned} $

Now, $\left(\mathbf{a}_2-\mathbf{a}_1\right)=(-2 \hat{\mathbf{i}}+\hat{\mathbf{k}}-2 \hat{\mathbf{i}}+\hat{\mathbf{j}})=-4 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$

$ \mathbf{b}_1 \times \mathbf{b}_2=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 0 & 2 \\ 1 & -1 & -1 \end{array}\right|=\hat{\mathbf{i}}(2)-\hat{\mathbf{j}}(-1-2)+\hat{\mathbf{k}}(-1)=2 \hat{\mathbf{i}}+3 \mathbf{j} \cdot \hat{\mathbf{i}} $

$ \begin{aligned} &\left|\mathbf{b}_1 \times \mathbf{b}_2\right|=\sqrt{2^2+3^2+1^2}=\sqrt{4+9+1}=\sqrt{14}\\ &\begin{aligned} \text { So, shortest distance } & =\left|\frac{(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}})(-4 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{14}}\right| \\ & =\left|\frac{-8+3-1}{\sqrt{14}}\right|=\frac{6}{\sqrt{14}}=\frac{3 \sqrt{2}}{\sqrt{7}} \end{aligned}\\ &\text { Hence, the shortest distance between skew lines is } \frac{3 \sqrt{2}}{\sqrt{7}} \end{aligned} $

To solve the problem of determining how the plane $x - y + z + 4 = 0$ divides the line segment connecting points $P(2, 3, -1)$ and $Q(1, 4, -2)$, we represent the division in the ratio $l:m$.

The coordinate of the point $R$ that divides the line segment from $P$ to $Q$ in the ratio $l:m$ is given by:

$ R\left(\frac{l \cdot 1 + m \cdot 2}{l + m}, \frac{l \cdot 4 + m \cdot 3}{l + m}, \frac{-2l - m}{l + m}\right) $

Since $R$ lies on the plane defined by the equation $x - y + z + 4 = 0$, substituting the coordinates of $R$ into the plane equation yields:

$ \left(\frac{l + 2m}{l + m}\right) - \left(\frac{4l + 3m}{l + m}\right) + \left(\frac{-2l - m}{l + m}\right) + 4 = 0 $

Simplifying this, we get:

$ \frac{l + 2m - 4l - 3m - 2l - m}{l + m} + 4 = 0 $

Simplifying further, we obtain:

$ \frac{-5l - 2m + 4l + 4m}{l + m} = 0 $

Which reduces to:

$ -l + 2m = 0 $

From this, it follows that:

$ l = 2m $

Therefore, the ratio $l:m$ is $2:1$, and consequently:

$ l + m = 2 + 1 = 3 $

To solve this problem, let's consider the given lines and their direction ratios:

The direction ratios of line $L_1$ are $(1, \alpha, \beta)$.

The direction ratios of line $L_2$ are $(-1, 2, 1)$.

The direction ratios of line $L_3$ are $(\alpha, 1, \beta)$.

Conditions Given:

Perpendicularity: The line $L_1$ is perpendicular to line $L_2$. Two lines are perpendicular if the dot product of their direction ratios equals zero.

$ 1 \times (-1) + \alpha \times 2 + \beta \times 1 = 0 $

This simplifies to:

$ -1 + 2\alpha + \beta = 0 \quad \text{(Equation 1)} $

Parallelism: The line $L_1$ is parallel to line $L_3$. Two lines are parallel if their direction ratios are proportional.

This implies:

$ \frac{1}{\alpha} = \frac{\alpha}{1} = \frac{\beta}{\beta} $

From the proportionality condition:

$ \alpha = 1 $

Solving Equations:

Using $\alpha = 1$ in Equation 1:

$ -1 + 2 \times 1 + \beta = 0 $

Simplifying:

$ -1 + 2 + \beta = 0 \quad \Rightarrow \quad \beta = -1 $

Thus, the values of $\alpha$ and $\beta$ are:

$ (\alpha, \beta) = (1, -1) $

We have,

$P\left(x_1, y_1, z_1\right)$ be the foot of perpendicular drawn from points $Q(2,-2,1)$ to the plane

$ x-2 y+z=1 $

Equation of $P Q$ is

$ \begin{aligned} & \frac{x_1-2}{1}=\frac{y_1-(-2)}{-2}=\frac{z_1-1}{1} \\ & \frac{x_1-2}{1}=\frac{y_1+2}{-2}=\frac{z_1-1}{1}=\lambda \end{aligned} $

General points on the line is

$ P(\lambda+2-2 \lambda-2, \lambda+1)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

If $P$ lies on the plane, then it will satisfy the equation

$ \begin{aligned} & \Rightarrow x_1-2 y_1+z_1-1=0 \\ & \Rightarrow(\lambda+2)+2(2 \lambda+2)+\lambda+1-1=0 \\ & \Rightarrow \lambda+2+4 \lambda+4+\lambda+1-1=0 \\ & \Rightarrow 6 \lambda+6=0 \Rightarrow \lambda=-1 \end{aligned} $

General point on the line is $P(1,0,0)$ If $d$ is the distance between the point $P$ and $Q$ then

$ \begin{aligned} d & =\left|\frac{a x_1+b y_1+c_1 z_1+d}{\sqrt{a^2+b^2+c^2}}\right| \\ \Rightarrow d & =\left|\frac{2+4+1-1}{\sqrt{1+4+1}}\right| \Rightarrow d=\left|\frac{6}{\sqrt{6}}\right| \end{aligned} $

Given, $l=x_1+y_1+z_1=1+0+0=1$

Then, $l+3 d^2=1+3\left(\frac{6}{\sqrt{6}}\right)^2=1+3 \times \frac{36}{6}$

$ l+3 d^2=19 $

Vector perpendicular to face $A B C$.

$ \begin{aligned} & =A B \times A C=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ 2 & -1 & 2 \end{array}\right| \\ & =\hat{\mathbf{i}}(2+1)-\hat{\mathbf{j}}(2-2)+\hat{\mathbf{k}}(-1-2) \\ & =3 \hat{\mathbf{i}}-0 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \end{aligned} $

Here, $A(1,2,1), B(2,3,2), C(3,1,3)$ and $D(2,1,3)$

$ \begin{aligned} & A B=O B-O A=(1,1,1) \\ & A C=(2,-1,2) \\ & A D=(1,-1,2) \text { and vector perpendicular to face } A B D \end{aligned} $

$ \begin{aligned} A B \times A D & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ 1 & -1 & 2 \end{array}\right| \\ & =\hat{\mathbf{i}}(2+1)-\hat{\mathbf{j}}(2-1)+\hat{\mathbf{k}}(-1-1)=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \end{aligned} $

Since, the angle between the faces $=$ Angle between their normals

Thus, $\cos \theta=\frac{9+0+6}{\sqrt{9+9} \sqrt{9+1+4}}$

$ \begin{aligned} & \cos \theta=\frac{15}{\sqrt{18} \sqrt{14}} \\ & \cos \theta=\frac{15}{\sqrt{9 \times 2} \sqrt{2 \times 7}} \\ & \cos \theta=\frac{15}{3 \sqrt{2} \sqrt{14}}=\frac{5}{\sqrt{2} \times \sqrt{2} \times \sqrt{7}} \\ & \cos \theta=\frac{5}{2 \sqrt{7}} \end{aligned} $

The tetrahedron is defined by the vertices $ A(3,2,4) $, $ B(x_1, y_1, 0) $, $ C(x_2, y_2, 0) $, and $ D(x_3, y_3, 0) $. The triangle $ \triangle BCD $ is formed by the lines $ y = x $, $ x + y = 6 $, and $ y = 1 $.

To find the centroid of the tetrahedron, we use the formula for the centroid of a tetrahedron with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, $(x_3, y_3, z_3)$, and $(x_4, y_4, z_4)$:

$ \left(\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4}\right) $

To determine the coordinates of the vertices $ B $, $ C $, and $ D $:

The intersection of the lines $ y = x $ and $ x + y = 6 $ gives the point $ (3, 3) $.

On the line $ y = x $, a point such as $ (5, 1) $ satisfies both conditions line $ x + y = 6 $.

A point on $ y = 1 $, such as $ (1, 1) $, also satisfies the line equation.

Thus, the coordinates of the vertices are:

$(x_2, y_2, z_2) = (3, 3, 0)$

$(x_3, y_3, z_3) = (5, 1, 0)$

$(x_4, y_4, z_4) = (1, 1, 0)$

Substituting these into the centroid formula gives:

$ \left(\frac{3 + 3 + 5 + 1}{4}, \frac{2 + 3 + 1 + 1}{4}, \frac{4 + 0 + 0 + 0}{4}\right) = \left(\frac{12}{4}, \frac{7}{4}, \frac{4}{4}\right) = \left(3, \frac{7}{4}, 1\right) $

Therefore, the centroid of the tetrahedron is $\left(3, \frac{7}{4}, 1\right)$.

Given, If $P(2, \beta, \alpha)$ lies on the plane $x+2 y-z-2=0$ and $Q(\alpha,-1, \beta)$ lies on the plane $2 x-y+3 z+6=0$ direction cosine of line $P Q=$ ?

Now, substitute the coordinates of $P$ into the plane equation $(x+2 y-z-2=0)$

$ 2+2 \beta-\alpha-2=0 \Rightarrow \alpha=2 \beta $

So, coordinate of $P$ are $(2, \beta, 2 \beta)$

Now, substitute the coordinates of $Q(\alpha,-1, \beta)$ into the plane equation

$ \begin{array}{r} (2 x-y+3 z+6=0) \\ 2 \alpha+1+3 \beta+6=0 \\ 2 \alpha+3 \beta+7=0 \end{array} $

Coordinate of $Q$ are $(\alpha,-1, \beta)$ direction ratios of the line $P Q$ are given by the differences in the coordinates of $Q$ and $P$.

$ P Q=(\alpha-2,-1-\beta, \beta-2 \beta) $

Here, $\alpha=-2$ and $\beta=-1$

From Eqs. (i) and (ii), we get

Magnitude of DR's

$ =\sqrt{(\alpha-2)^2+(-1-\beta)^2+(\beta-2 \beta)^2} $

Direction cosine of are $(l, m, n)$

$ \begin{aligned} & l=\frac{\alpha-2}{\sqrt{(\alpha-2)^2+(-1-\beta)^2+(\beta-2 \beta)^2}} \\ & m=\frac{-1-\beta}{\sqrt{(\alpha-2)^2+(-1-\beta)^2+(\beta-2 \beta)^2}} \\ & n=\frac{\beta-2 \beta}{\sqrt{(\alpha-2)^2+(-1-\beta)^2+(\beta-2 \beta)^2}} \\ & \text { So, } l=\frac{-2-2}{\sqrt{(-2-2)^2+(-1+1)^2+(1)^2}} \\ & \quad=\frac{-4}{\sqrt{17}} \\ & m=\frac{-1+1}{\sqrt{17}}=0 \\ & n=\frac{-1+2}{\sqrt{17}}=\frac{1}{\sqrt{17}}:\left(\frac{-4}{\sqrt{17}}, 0, \frac{1}{\sqrt{17}}\right) \end{aligned} $

To find the foot of the perpendicular from the point $ (1, 2, 1) $ to the plane $ \pi $, which passes through the point $(-2, 1, -1)$ and is parallel to the plane $2x - y + 2z = 0$, we can follow these steps:

Step 1: Determine the Equation of the Plane $\pi$

The plane $\pi$ has the same normal vector as the plane $2x - y + 2z = 0$, which is $2\hat{\mathbf{i}} - \hat{\mathbf{j}} + 2\hat{\mathbf{k}}$.

So, the general equation of the plane $\pi$ will be:

$ 2x - y + 2z = d $

Given that the plane passes through the point $(-2, 1, -1)$, we substitute these coordinates into the equation:

$ 2(-2) - 1 + 2(-1) = d $

$ -4 - 1 - 2 = d $

$ d = -7 $

Thus, the equation of the plane $\pi$ is:

$ 2x - y + 2z = -7 $

Step 2: Find the Foot of the Perpendicular

Let $ P(a, b, c) $ be the foot of the perpendicular from the point $ Q(1, 2, 1) $ to the plane. The vector from $ Q $ to $ P $, $ \overrightarrow{QP} $, must be parallel to the normal vector of the plane, which is given by:

$ 2\hat{\mathbf{i}} - \hat{\mathbf{j}} + 2\hat{\mathbf{k}} $

Thus, we can express:

$ \overrightarrow{QP} = (a - 1)\hat{\mathbf{i}} + (b - 2)\hat{\mathbf{j}} + (c - 1)\hat{\mathbf{k}} $

Equating components, we get the parallel condition:

$ \frac{a - 1}{2} = \frac{b - 2}{-1} = \frac{c - 1}{2} = \lambda $

From here, solve for $ a $, $ b $, and $ c $:

$ a = 2\lambda + 1 $

$ b = 2 - \lambda $

$ c = 2\lambda + 1 $

Step 3: Use the Plane Equation

Since point $ P(a, b, c) $ lies on the plane:

$ 2(2\lambda + 1) - (2 - \lambda) + 2(2\lambda + 1) = -7 $

$ 4\lambda + 2 - 2 + \lambda + 4\lambda + 2 = -7 $

$ 9\lambda + 2 = -7 $

Solving for $\lambda$:

$ 9\lambda = -9 $

$ \lambda = -1 $

Step 4: Substitute to Find Coordinates

Substituting $\lambda = -1$ back into the equations for $a$, $b$, and $c$:

$ a = 2(-1) + 1 = -1 $

$ b = 2 - (-1) = 3 $

$ c = 2(-1) + 1 = -1 $

Thus, the coordinates of the foot of the perpendicular are:

$ P(-1, 3, -1) $

Given the direction ratios $(2, 5, 1)$ of a line and the equation of the plane $8x + 2y - z = 14$, we aim to find the angle between the line and the plane.

To determine the angle $\theta$ between the line and the plane, we use the formula for the sine of the angle:

$ \sin \theta = \frac{a_1b_1 + a_2b_2 + a_3b_3}{\sqrt{a_1^2 + a_2^2 + a_3^2} \cdot \sqrt{b_1^2 + b_2^2 + b_3^2}} $

Here, the direction ratios of the line are $(a_1, a_2, a_3) = (2, 5, 1)$ and the normal to the plane, derived from the coefficients of $x$, $y$, and $z$, is $(b_1, b_2, b_3) = (8, 2, -1)$.

Substituting these into the formula, we calculate:

$ \sin \theta = \frac{2 \times 8 + 5 \times 2 + 1 \times (-1)}{\sqrt{2^2 + 5^2 + 1^2} \cdot \sqrt{8^2 + 2^2 + (-1)^2}} $

Simplifying, we get:

$ = \frac{16 + 10 - 1}{\sqrt{4 + 25 + 1} \cdot \sqrt{64 + 4 + 1}} $

$ = \frac{25}{\sqrt{30} \cdot \sqrt{69}} $

This simplifies further to:

$ = \frac{25}{3 \sqrt{230}} = \frac{25}{\sqrt{2070}} $

Therefore, the angle $\theta$ is given by:

$ \theta = \sin^{-1}\left(\frac{25}{3 \sqrt{230}}\right) = \sin^{-1}\left(\frac{25}{\sqrt{2070}}\right) $

To find the direction cosines of the line of intersection of the planes, we start with the given equations:

$ x + 2y + z - 4 = 0 $

$ 2x - y + z - 3 = 0 $

The direction ratios for these planes are derived from their coefficients, so for equation (i), the direction ratios are $ (1, 2, 1) $ and for equation (ii), they are $ (2, -1, 1) $.

To find the direction ratios of the line of intersection, we calculate the cross product of these direction ratios using:

$ \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 1 \\ 2 & -1 & 1 \end{vmatrix} $

Calculating the cross product:

$ \begin{aligned} &= (2 \cdot 1 + 1 \cdot 1) \hat{\mathbf{i}} - (1 \cdot 1 - 2 \cdot 1) \hat{\mathbf{j}} + (1 \cdot (-1) - 2 \cdot 2) \hat{\mathbf{k}} \\ &= (2 + 1) \hat{\mathbf{i}} - (1 - 2) \hat{\mathbf{j}} + (-1 - 4) \hat{\mathbf{k}} \\ &= 3 \hat{\mathbf{i}} + \hat{\mathbf{j}} - 5 \hat{\mathbf{k}} \end{aligned} $

Thus, the direction ratios are $ \langle 3, 1, -5 \rangle $.

To convert these into direction cosines, we normalize the vector:

$ \left\langle \frac{3}{\sqrt{9+1+25}}, \frac{1}{\sqrt{9+1+25}}, \frac{-5}{\sqrt{9+1+25}} \right\rangle = \left\langle \frac{3}{\sqrt{35}}, \frac{1}{\sqrt{35}}, \frac{-5}{\sqrt{35}} \right\rangle $

These are the direction cosines of the line of intersection.

Given the direction ratios for lines $ L_1 $ and $ L_2 $ as $(3, 1, -5)$ and $(2, 3, -1)$ respectively, we need to find the equation of the plane that contains both lines $ L_1 $ and $ L_2 $.

To derive the equation of the plane, we use the determinant method with the direction ratios of the lines as follows:

$ \left|\begin{array}{ccc} x & y & z \\ 3 & 1 & -5 \\ 2 & 3 & -1 \\ \end{array}\right| = 0 $

This expands to:

$ x((-1 \times 3) - (15 \times 1)) - y((-3 \times 1) - (10 \times 3)) + z((9 \times 3) - (2 \times (-1))) $

After simplifying, we obtain:

$ x(-1 + 15) - y(-3 + 10) + z(9 - 2) = 0 $

Which further simplifies to:

$ 14x - 7y + 7z = 0 $

Dividing through by 7, we get the simplified equation of the plane:

$ 2x - y + z = 0 $

To find the position vector of point $ R $, where a line passing through points $ A $ and $ B $ intersects the plane defined by points $ O $, $ C $, and $ D $, we start by calculating the vector equations and the intersection.

Position vectors of the points are:

$ A = \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + \hat{\mathbf{k}} $

$ B = -2 \hat{\mathbf{i}} + 3 \hat{\mathbf{k}} $

$ C = 2 \hat{\mathbf{i}} + \hat{\mathbf{j}} $

$ D = 4 \hat{\mathbf{k}} $

Step 1: Find the vector $\mathbf{AB}$:

$ \mathbf{AB} = B - A = (-2 \hat{\mathbf{i}} + 3 \hat{\mathbf{k}}) - (\hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + \hat{\mathbf{k}}) = -3 \hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}} $

Step 2: Parametric equation of the line through $ A $ and $ B $:

$ \mathbf{r} = \mathbf{A} + t \mathbf{AB} = (\hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + \hat{\mathbf{k}}) + t (-3 \hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}}) $

Simplifying, we get:

$ \mathbf{r} = (1 - 3t) \hat{\mathbf{i}} + (2 - 2t) \hat{\mathbf{j}} + (1 + 2t) \hat{\mathbf{k}} $

Step 3: Equation for the plane through $ O $, $ C $, and $ D $:

$\mathbf{OC} = C - O = 2 \hat{\mathbf{i}} + \hat{\mathbf{j}}$

$\mathbf{OD} = D - O = 4 \hat{\mathbf{k}}$

Calculate the normal vector $\mathbf{n}$:

$ \mathbf{n} = \mathbf{OC} \times \mathbf{OD} = \left| \begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & 0 \\ 0 & 0 & 4 \end{array} \right| = 4 \hat{\mathbf{i}} - 8 \hat{\mathbf{j}} $

Step 4: Plane equation based on $\mathbf{n}$:

The plane equation is:

$ 4x - 8y = 0 $

Step 5: Substitute the parametric line equation into the plane equation:

$ 4(1 - 3t) - 8(2 - 2t) = 0 $

Simplify the equation:

$ 4 - 12t - 16 + 16t = 0 \\ 4t - 12 = 0 \\ t = 3 $

Step 6: Calculate the coordinates of $ R $:

Substitute $ t = 3 $ into $\mathbf{r} = (1 - 3t) \hat{\mathbf{i}} + (2 - 2t) \hat{\mathbf{j}} + (1 + 2t) \hat{\mathbf{k}}$:

$ x = 1 - 3(3) = -8 $

$ y = 2 - 2(3) = -4 $

$ z = 1 + 2(3) = 7 $

Thus, the position vector of $ R $ is:

$ -8 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} + 7 \hat{\mathbf{k}} $

The distance of the point $ O(\mathbf{O}) $ from the plane, given by the equation $\mathbf{r} \cdot (\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) = 5$, is measured along the direction of the vector $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}$.

First, we identify the plane's equation:

$ x + y + z = 5 $

The line parallel to the vector $2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} - 6 \hat{\mathbf{k}}$ can be represented parametrically as:

$ \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} = K $

This gives us:

$ x = 2K, \quad y = 3K, \quad z = -6K $

For the point $(2K, 3K, -6K)$ to be on the plane, it must satisfy the plane equation:

$ 2K + 3K - 6K = 5 $

Solving this, we find:

$ -K = 5 \quad \Rightarrow \quad K = -5 $

Thus, the coordinates of the point on the line are:

$ (-10, -15, 30) $

Now, to find the distance of this point from the origin, we calculate the Euclidean distance:

$ \begin{aligned} \text{Distance} &= \sqrt{(-10)^2 + (-15)^2 + 30^2} \\ &= \sqrt{100 + 225 + 900} \\ &= \sqrt{1225} \\ &= 35 \\ \end{aligned} $

Therefore, the required distance is 35.

To find the point of intersection $ P(a, b, c) $ of the lines $ AB $ and $ CD $, let's first determine the equations of these lines.

Equation of Line $ AB $

For line $ AB $, given points $ A(1,0,2) $ and $ B(2,1,0) $, the direction ratios are:

$ (2-1, 1-0, 0-2) = (1, 1, -2) $

Thus, the equation of line $ AB $ is:

$ \frac{x-1}{1} = \frac{y-0}{1} = \frac{z-2}{-2} $

Equation of Line $ CD $

For line $ CD $, given points $ C(2,-5,3) $ and $ D(0,3,2) $, the direction ratios are:

$ (0-2, 3+5, 2-3) = (-2, 8, -1) $

Thus, the equation of line $ CD $ is:

$ \frac{x}{-2} = \frac{y-3}{8} = \frac{z-2}{-1} = K $

Therefore, in parametric form, the point $ P $ on line $ CD $ can be expressed as:

$ (x, y, z) = (2K, 3 - 8K, K + 2) $

Point $ P $ on Line $ AB $

Since point $ P $ also lies on line $ AB $, we equate the expressions:

$ 2K - 1 = 3 - 8K = \frac{K}{-2} $

Solving the first two equations:

$ 2K - 1 = 3 - 8K $

$ 10K = 4 $

$ K = \frac{2}{5} $

Substituting $ K = \frac{2}{5} $ into the parametric form:

$ x = 2K = 2 \times \frac{2}{5} = \frac{4}{5} $

$ y = 3 - 8K = 3 - 8 \times \frac{2}{5} = -\frac{1}{5} $

$ z = K + 2 = \frac{2}{5} + 2 = \frac{12}{5} $

Thus, the coordinates of $ P $ are:

$ \left(\frac{4}{5}, -\frac{1}{5}, \frac{12}{5}\right) $

Therefore, the sum $ a + b + c $ is:

$ a + b + c = \frac{4}{5} - \frac{1}{5} + \frac{12}{5} = 3 $

We start with the relations for the direction cosines of two lines:

$ l+m-n=0 \quad \text{and} \quad lm - 2mn + nl = 0 $

From $ l+m-n=0 $, we can express $ n $ as:

$ n = l + m $

Substituting this into the second equation, we have:

$ lm - 2m(l+m) + (l+m)l = 0 $

This simplifies to:

$ lm - 2lm - 2m^2 + l^2 + lm = 0 $

$ l^2 = 2m^2 $

Thus, $ l = \pm \sqrt{2}m $.

Now, we express $ n $ based on these values of $ l $:

$ n = \pm \sqrt{2}m + m = (\pm \sqrt{2} + 1)m $

We deduce two scenarios for the direction cosines:

$ l:m:n = \sqrt{2}:1:\sqrt{2}+1 $

$ l:m:n = -\sqrt{2}:1:-\sqrt{2}+1 $

The cosine of the acute angle $ \theta $ between the lines is given by:

$ \cos \theta = \frac{\sqrt{2} \times (-\sqrt{2}) + 1 \times 1 + (\sqrt{2} + 1)(-\sqrt{2} + 1)}{\sqrt{2+1+2+1+2\sqrt{2}} \sqrt{2+1+2+1-2\sqrt{2}}} $

This simplifies as:

$ = \frac{|-2 + 1 - 1|}{\sqrt{6 + 2\sqrt{2}} \sqrt{6 - 2\sqrt{2}}} = \frac{2}{\sqrt{36 - 8}} = \frac{1}{\sqrt{7}} $

Therefore, the cosine of the acute angle $ \theta $ is $\frac{1}{\sqrt{7}}$.

To determine the locus of point $ P $, consider a plane $\pi$ intersecting the coordinate axes at points $ A $, $ B $, and $ C $.

The equation of the plane is given by:

$ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 $

where the plane intersects the $ X $-axis at $ A(a, 0, 0) $, $ Y $-axis at $ B(0, b, 0) $ and $ Z $-axis at $ C(0, 0, c) $.

The distance from a fixed point $(1, 1, 1)$ to this plane is 12 units. The formula for the distance $ d $ from a point $(x_1, y_1, z_1)$ to a plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ is given by:

$ d = \frac{\left| \frac{x_1}{a} + \frac{y_1}{b} + \frac{z_1}{c} - 1 \right|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} $

Substituting the point $(1, 1, 1)$ into the distance formula, we have:

$ \frac{\left| \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 1 \right|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 12 $

Squaring both sides, we obtain:

$ \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 1\right)^2 = 144 \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right) $

Thus, we identify that point $ P $, which is the intersection of planes through points $ A $, $ B $, $ C $ and parallel to the coordinate planes, adheres to the following equation:

$ \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} - 1\right)^2 = 144 \left(\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}\right) $

Therefore, the locus of $ P $ satisfies this equation.

To find the shortest distance between two skew lines given by:

$\mathbf{r} = (-\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 3\hat{\mathbf{k}}) + t(3\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 2\hat{\mathbf{k}})$

$\mathbf{r} = (7\hat{\mathbf{i}} + 4\hat{\mathbf{k}}) + s(\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + 2\hat{\mathbf{k}})$

follow these steps:

Define the Position Vectors and Direction Vectors:

$\mathbf{a}_1 = -\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 3\hat{\mathbf{k}}$

$\mathbf{b}_1 = 3\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 2\hat{\mathbf{k}}$

$\mathbf{a}_2 = 7\hat{\mathbf{i}} + 4\hat{\mathbf{k}}$

$\mathbf{b}_2 = \hat{\mathbf{i}} - 2\hat{\mathbf{j}} + 2\hat{\mathbf{k}}$

Calculate the Vector Connecting Points on the Lines:

$ \mathbf{a}_2 - \mathbf{a}_1 = (7 + 1)\hat{\mathbf{i}} + (0 + 2)\hat{\mathbf{j}} + (4 + 3)\hat{\mathbf{k}} = 8\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 7\hat{\mathbf{k}} $

Find the Cross Product of the Direction Vectors:

$ \mathbf{b}_1 \times \mathbf{b}_2 = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & -2 & -2 \\ 1 & -2 & 2 \end{vmatrix} $

$ = \hat{\mathbf{i}}((-2) \cdot 2 - (-2) \cdot (-2)) - \hat{\mathbf{j}}(3 \cdot 2 - (-2) \cdot 1) + \hat{\mathbf{k}}(3 \cdot (-2) - (-2) \cdot 1) $

$ = \hat{\mathbf{i}}(-4 - 4) - \hat{\mathbf{j}}(6 + 2) + \hat{\mathbf{k}}(-6 + 2) $

$ = -8\hat{\mathbf{i}} - 8\hat{\mathbf{j}} - 4\hat{\mathbf{k}} $

Calculate the Magnitude of the Cross Product:

$ \left|\mathbf{b}_1 \times \mathbf{b}_2\right| = \sqrt{(-8)^2 + (-8)^2 + (-4)^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12 $

Find the Required Distance:

$ \text{Distance} = \frac{|(8\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 7\hat{\mathbf{k}}) \cdot (-8\hat{\mathbf{i}} - 8\hat{\mathbf{j}} - 4\hat{\mathbf{k}})|}{12} $

$ = \frac{|8(-8) + 2(-8) + 7(-4)|}{12} $

$ = \frac{|-64 - 16 - 28|}{12} $

$ = \frac{108}{12} = 9 $

Thus, the shortest distance between these skew lines is 9.