Three Dimensional Geometry

Step 1: Find the Direction Vector of the Line

The line joins the points $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$ (which is (1, 2, 0)) and $\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ (which is (0, 1, -2)).

So, the direction vector is found by subtracting these: $(0-1, 1-2, -2-0) = (-1, -1, -2)$.

Step 2: Write the Equation of the Line

The equation of the line is: $(x, y, z) = (1, 2, 0) + t(-1, -1, -2)$.

This gives: $x = 1-t$, $y = 2-t$, $z = -2t$.

Step 3: Find the Equation of the Plane

The plane goes through $2\hat{\mathbf{i}}-\hat{\mathbf{j}}$ (2, -1, 0), $2\hat{\mathbf{j}}+3\hat{\mathbf{k}}$ (0, 2, 3), and $\hat{\mathbf{k}}-2\hat{\mathbf{i}}$ (-2, 0, 1).

Step 4: Get Two Vectors in the Plane

Let $P = (2, -1, 0)$, $Q = (0, 2, 3)$, $R = (-2, 0, 1)$. Compute: $PQ = Q-P = (-2, 3, 3)$, $PR = R-P = (-4, 1, 1)$.

Step 5: Find the Normal Vector by Cross Product

The normal vector of the plane is $PQ \times PR$:

$ \begin{aligned} & = \left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -2 & 3 & 3 \\ -4 & 1 & 1 \end{array}\right| \\ & = \hat{\mathbf{i}}(3\cdot1 - 3\cdot1) - \hat{\mathbf{j}}(-2\cdot1 - 3\cdot-4) + \hat{\mathbf{k}}(-2\cdot1 - 3\cdot-4) \\ & = \hat{\mathbf{i}}(0) - \hat{\mathbf{j}}(10) + \hat{\mathbf{k}}(10) \\ & = -10 \hat{\mathbf{j}} + 10 \hat{\mathbf{k}} \end{aligned} $

Step 6: Write Equation of the Plane

The normal vector is $(0, -10, 10)$. The plane passes through $P = (2, -1, 0)$, so the equation is: $0(x-2) - 10(y+1) + 10(z-0) = 0$.

This simplifies to $-10(y+1) + 10z = 0$, or $z - y = 1$.

Step 7: Find Where the Line Meets the Plane

Plug the line equations $y=2-t$, $z=-2t$ into the plane equation $z-y=1$:

$-2t - (2-t) = 1$

$-2t - 2 + t = 1$

$-t - 2 = 1$

$-t = 3$

$t = -3$

Step 8: Find the Intersection Point

Put $t=-3$ into the line equations:

$x = 1 - (-3) = 4$

$y = 2 - (-3) = 5$

$z = -2 \times (-3) = 6$

So, the point $\mathbf{r} = (4, 5, 6)$.

Step 9: Find $\mathbf{r} \cdot (\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})$

This is $4 \times 1 + 5 \times 1 + 6 \times 1 = 15$.

A plane passing through the line of intersection of two planes

$ \begin{aligned} & \therefore \mathbf{r} \cdot((\hat{\mathbf{i}}-2 \hat{\mathbf{k}})+\lambda(2 \hat{\mathbf{j}}+\hat{\mathbf{k}}))=3+5 \lambda \\ & \Rightarrow \mathbf{r} \cdot(\hat{\mathbf{i}}+2 \lambda \hat{\mathbf{j}}+(-2+\lambda) \hat{\mathbf{k}})=3+5 \lambda \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{aligned} &\begin{aligned} & \text { put } \mathbf{r}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \text { then, } 1+4 \lambda+3(-2+\lambda)=3+5 \lambda \\ & \Rightarrow \quad 1-6+7 \lambda=3+5 \lambda \Rightarrow \lambda=4 \end{aligned}\\ &\text { Substitute } \lambda \text { in Eq. (i) }\\ &\mathbf{r} \cdot(\hat{\mathbf{i}}+8 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})=23 \end{aligned} $

$a=(-1,2,3), B=(2,-3,1)$,

$ \begin{aligned} & C=(3,1,-2) \\ & A B=\sqrt{(2+1)^2+(-3-2)^2+(1-3)^2}=\sqrt{37} \\ & B C=\sqrt{(3-2)^2+(1+3)^2+(-2-1)^2}=\sqrt{26} \end{aligned} $

$ A C=\sqrt{(3+1)^2+(1-2)^2+(-2-3)^2}=\sqrt{42} $

$\because A B \neq B C \neq A C$ and it does not follow right angled triangle property.

Hence, $A, B, C$ form a scalene triangle.

$\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \cos \frac{\pi}{3}=\frac{1}{2}$

and we know, $l^2+m^2+n^2=1$

$ \begin{aligned} & \Rightarrow\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{2}\right)^2+\cos ^2 \theta=1 \\ & \Rightarrow \frac{1}{2}+\frac{1}{4}+\cos ^2 \theta=1 \Rightarrow \cos ^2 \theta=\frac{1}{4} \\ & \Rightarrow \cos \theta=\sqrt{\frac{1}{4}}=\frac{1}{2} \end{aligned} $

Hence, direction cosines are $\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}$

Direction ratio of normal to the plane $2 x-3 y+5 z=7$ is $2,-3,5$ and to $5 x+2 y-3 z=11$ is $\langle 5,2,-3\rangle \because$ Required plane is perpendicular to both, its normal vector $\langle 1, b, c\rangle$ must be perpendicular to both given normal vectors.

$ \therefore\langle 1, b, c\rangle \cdot\langle 2,-3,5\rangle=2-3 b+5 c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $\langle 1, b, c\rangle \cdot\langle 5,2,-3\rangle$

$ =5+2 b-3 c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

By Eq. (i) and (ii), we get

$ b=-31, c=-19 $

The equation of plane is $x-31 y-19 z+d=0$ is passes through $(3,2,5)$

$ \therefore 3-62-95+d=0 \Rightarrow d=154 $

Hence, $2 b+3 c+d=2(-31) +3(-19)+154=35$

Let $A(2,-1,1), B(1,-3,-5)$ and $C(3,-4,-4)$ be vertices of $\triangle A B C$

$ \begin{aligned} \therefore \quad a & =\sqrt{4+1+1}=\sqrt{6} \\ b & =\sqrt{1+9+25}=\sqrt{35} \\ c & =\sqrt{1+4+36}=\sqrt{41} \\ \because a^2+b^2 & =c^2 \end{aligned} $

$\therefore \triangle A B C$ is right-angled triangle

∴ Circumradius $=\frac{c}{2}=\frac{\sqrt{41}}{2}$

∵ Median $A D$ is equally inclined with coordinate axes

$ A D=\frac{\lambda-5}{2} \hat{\mathbf{i}}+\hat{\mathbf{j}}+\left(\frac{\mu-8}{2}\right) \hat{\mathbf{k}} $

Let $\alpha, \beta, \gamma$ be direction ratio of $A D \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$

$ \begin{array}{rlrl} & \because & \alpha & =\beta=\gamma \\ & & 3 \cos ^2 \alpha & =1 \Rightarrow \cos ^2 \alpha=\frac{1}{3} \\ & \therefore & \frac{\lambda-5}{2} & =1, \frac{\mu-8}{2}=1 \\ & & \lambda=7, \mu=10 \\ & \therefore & 10 \lambda-7 \mu & =0 \end{array} $

∵ Plane is $\perp^r$ to planes $x+2 y-z=1$ and $3 x-4 y+z=5$

$ \begin{aligned} \therefore \mathbf{n} & =\mathbf{n}_1 \times \mathbf{n}_2 \\ & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & -1 \\ 3 & -4 & 1 \end{array}\right|=-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-10 \hat{\mathbf{k}} \end{aligned} $

∴ Equation of plane passing through origin and $\perp^r$ to $-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}$

$ \begin{aligned} \mathbf{r} \cdot \mathbf{n} & =\mathbf{a} \cdot \mathbf{n} \\ \Rightarrow \quad-2 x-4 y-10 z & =0 \\ \Rightarrow \quad x+2 y+5 z & =0 \end{aligned} $

$\because L_1$ passes through $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ i.e., $(1,1,0)$ and $k-i$ i.e., $(-1,0,1)$

∴ Equation of $L_1$

$ \begin{aligned} & \frac{x-1}{-1-1}=\frac{y-1}{0-1}=\frac{z-0}{1-0} \\ \Rightarrow & \frac{x-1}{-2}=\frac{y-1}{-1}=\frac{z}{1} \\ \Rightarrow & \frac{x-1}{2}=\frac{y-1}{1}=\frac{z}{-1} \end{aligned} $

any point ( $2 \lambda+1, \lambda+1,-\lambda$ )

Similarly $L_2$ passes through $(0,1,2)$ and parallel $\mathbf{i}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$

⇒ Equation of $L_2$

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

Any point on $L_2 \Rightarrow(\mu, \mu+1, \mu+2)$

∴ Point of intersection

$ \begin{aligned} & 2 \lambda+1=\mu, \lambda+1=\mu+1,-\lambda=\mu+2 \\ & \therefore \quad \lambda=\mu \Rightarrow \lambda=\mu=-1 \end{aligned} $

∴ Point of intersection $=(-1,0,1)$

$ \therefore y-x=1=z $

Let $P(x, y, z)$ be a point that is the same distance from each of the points $A(2,0,3)$, $B(0,3,2)$, and $C(0,0,1)$.

To find $P$, let's set all the distances equal:

Step 1: Equal Distance from $A$ and $B$

The distance from $P$ to $A$ is $\sqrt{(x-2)^2 + (y-0)^2 + (z-3)^2}$.

The distance from $P$ to $B$ is $\sqrt{(x-0)^2 + (y-3)^2 + (z-2)^2}$.

Set these distances equal (we can remove the square roots since both must be positive):

$(x-2)^2 + y^2 + (z-3)^2 = x^2 + (y-3)^2 + (z-2)^2$

Simplify Equation 1:

Expand both sides:

Left: $x^2 - 4x + 4 + y^2 + z^2 - 6z + 9$
Right: $x^2 + y^2 - 6y + 9 + z^2 - 4z + 4$

Subtract the right side from the left side:

$-4x - 6z + 4 + 9 - (-6y + 9 - 4z + 4) = 0$

$-4x - 6z + 13 + 6y - 9 + 4z - 4 = 0$

Group like terms:

$-4x + 6y - 2z = 0$

Or, $2x - 3y + z = 0$ (divide by -2)

This is Equation (i).

Step 2: Equal Distance from $B$ and $C$

Write their distances equal:

$(x-0)^2 + (y-3)^2 + (z-2)^2 = x^2 + y^2 + (z-1)^2$

Expand both sides:

Left: $x^2 + y^2 - 6y + 9 + z^2 - 4z + 4$
Right: $x^2 + y^2 + z^2 - 2z + 1$

Subtract the right side from the left side:

$-6y + 9 - 4z + 4 - (-2z + 1) = 0$

$-6y + 13 - 4z + 2z - 1 = 0$

$-6y - 2z + 12 = 0$

Or, $3y + z = 6$ (divide by -2)

This is Equation (ii).

Step 3: Solve Both Equations Together

We now solve these:

Equation (i): $2x - 3y + z = 0$
Equation (ii): $3y + z = 6$

From Equation (ii): $z = 6 - 3y$

Substitute $z$ in Equation (i):
$2x - 3y + (6 - 3y) = 0$
$2x - 6y + 6 = 0$

$2x = 6y - 6$
$x = 3y - 3$

Let's set $y = 2$ (for a possible solution):
$\to x = 3(2) - 3 = 6 - 3 = 3$
$z = 6 - 3(2) = 6 - 6 = 0$

So, the point $P$ with coordinates $(3, 2, 0)$ fits both equations.

We are given the direction ratios for $L_1$ as $1, -2, 2$.

We are also given the direction ratios for $L_2$ as $-2, 3, -6$.

To find the direction ratios for a line that is perpendicular to both $L_1$ and $L_2$, we need to take the cross product of the direction ratios of $L_1$ and $L_2$.

Let us use the formula for the cross product. If the direction ratios of the first line are $(a_1, b_1, c_1)$ and of the second line are $(a_2, b_2, c_2)$, then the perpendicular line will have direction ratios:

$(b_1c_2 - c_1b_2,\; c_1a_2 - a_1c_2,\; a_1b_2 - b_1a_2)$

Now substitute the given values:

$( -2 \times -6 - 2 \times 3,\; 2 \times -2 - 1 \times -6,\; 1 \times 3 - ( -2 ) \times -2 )$

This becomes:

$(12 - 6,\; -4 + 6,\; 3 - 4 )$

Which simplifies to:

$6, 2, -1$

So, the direction ratios for the required line are $6, 2, -1$.

Step 1: Formula for the image of a point with respect to a plane

If you have a point $A(x_1, y_1, z_1)$ and a plane $ax + by + cz + d = 0$, the image point $B(\alpha, \beta, \gamma)$ is found using the following:

$ \frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{\gamma - z_1}{c} = \frac{-2(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2} $

Step 2: Plug in the values

Here, $A(1,1,1)$ and the plane is $4x+2y+4z+1=0$. So, $a=4$, $b=2$, $c=4$, and $d=1$.

Calculate $ax_1 + by_1 + cz_1 + d = 4 \times 1 + 2 \times 1 + 4 \times 1 + 1 = 4 + 2 + 4 + 1 = 11$.

Calculate $a^2 + b^2 + c^2 = 4^2 + 2^2 + 4^2 = 16 + 4 + 16 = 36$.

Now, substitute both results into the formula:

$ \frac{\alpha - 1}{4} = \frac{\beta - 1}{2} = \frac{\gamma - 1}{4} = \frac{-2 \times 11}{36} $

This simplifies to:

$ \frac{\alpha - 1}{4} = \frac{\beta - 1}{2} = \frac{\gamma - 1}{4} = \frac{-22}{36} = \frac{-11}{18} $

Step 3: Solve for the coordinates

For $\alpha:$ $ \frac{\alpha - 1}{4} = \frac{-11}{18} \Rightarrow \alpha-1 = 4 \cdot \frac{-11}{18} = \frac{-44}{18} = \frac{-22}{9} \Rightarrow \alpha = 1 + \frac{-22}{9} = \frac{-13}{9} $

For $\beta:$ $ \frac{\beta - 1}{2} = \frac{-11}{18} \Rightarrow \beta-1 = 2 \cdot \frac{-11}{18} = \frac{-22}{18} = \frac{-11}{9} \Rightarrow \beta = 1 + \frac{-11}{9} = \frac{-2}{9} $

For $\gamma:$ $ \frac{\gamma - 1}{4} = \frac{-11}{18} \Rightarrow \gamma-1 = 4 \cdot \frac{-11}{18} = \frac{-44}{18} = \frac{-22}{9} \Rightarrow \gamma = 1 + \frac{-22}{9} = \frac{-13}{9} $

Step 4: Add the coordinates

$ \alpha + \beta + \gamma = \frac{-13}{9} + \frac{-2}{9} + \frac{-13}{9} = \frac{-13-2-13}{9} = \frac{-28}{9} $

Assertion Given lines $\mathbf{r}=\mathbf{a}+t \mathbf{b}$ and $\mathbf{r}=\mathbf{p}+s \mathbf{q}$

We know that the lines are coplanar if $(\mathbf{a}-\mathbf{p}) \cdot(\mathbf{b} \times \mathbf{q})=0$

Thus, the assertion is wrong.

Reason The shortest distance between two skew lines is

$ \begin{aligned} & \qquad d=\frac{|(\mathbf{a}-\mathbf{p}) \cdot(\mathbf{b} \times \mathbf{q})|}{|\mathbf{b} \times \mathbf{q}|} \\ & \Rightarrow \quad d|\mathbf{b} \times \mathbf{q}|=|(\mathbf{a}-\mathbf{p}) \cdot(\mathbf{b} \times \mathbf{q})| \\ & \text { Thus, the reason is } \end{aligned} $

So, $A$ is false, but $R$ is true.

Let the two fixed points be $A(-3,1,2)$ and $B(1,-2,4)$.

Let $P(x, y, z)$ be the point at which the line segment $A B$ subtends a right angle.

Then, $\angle A P B=90^{\circ}$

$ \Rightarrow \quad \mathbf{P A} \cdot \mathbf{P B}=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Now, $\mathbf{P A}=(-3-x, 1-y, 2-z)$

$ \mathbf{P B}=(1-x,-2-y, 4-z) $

So, $\mathbf{P A} \cdot \mathbf{P B}=(-3-x)(1-x) +(1-y)(-2-y)+(2-z)(4-z)=0 $

$ \begin{array}{cc} \Rightarrow & -3+3 x-x+x^2-2-y+2 y +y^2+8-2 z-4 z+z^2=0 \\ \Rightarrow & x^2+2 x+y^2+y+z^2-6 z+3=0 \\ \Rightarrow & x^2+y^2+z^2+2 x+y-6 z+3=0 \end{array} $

Given, vertices of $\triangle A B C$ are $A(1,2,3), B(2,3,-1)$ and $C(3,-1,-2)$

Now, $A B=B-A=(2,3,-1) -(1,2,3)=(1,1,-4) $

Now,

$ \mathbf{C B}=(3,-1,-2)-(2,3,-1)=(1,-4,-1) $

$ \begin{aligned}So,\,\, |\mathbf{A B}| & =\sqrt{1^2+1^2+(-4)^2} \\ & =\sqrt{1+1+16}=\sqrt{18}=3 \sqrt{2} \\ |\mathbf{C B}| & =\sqrt{1^2+(-4)^2+(-1)^2} \\ & =\sqrt{18}=3 \sqrt{2} \end{aligned} $

So, d.c. of $\mathbf{A B}=\frac{\mathbf{A B}}{|\mathbf{A B}|}=\frac{(1,1,-4)}{3 \sqrt{2}}$

$ =\left(\frac{1}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}, \frac{-4}{3 \sqrt{2}}\right) $

And d.c. of $\mathbf{C B}=\frac{\mathbf{C B}}{|\mathbf{C B}|}=\frac{(1,-4,-1)}{3 \sqrt{2}}$

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left(\frac{1}{3 \sqrt{2}}, \frac{-4}{3 \sqrt{2}}, \frac{-1}{3 \sqrt{2}}\right) $

$\therefore$ d.c.'s of the internal bisector of $\angle A B C$ are

$ \begin{aligned} & \left(\frac{1}{3 \sqrt{2}}+\frac{1}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}-\frac{4}{3 \sqrt{2}}, \frac{-4}{3 \sqrt{2}}-\frac{1}{3 \sqrt{2}}\right) \\ & =\left(\frac{2}{3 \sqrt{2}}, \frac{-3}{3 \sqrt{2}}, \frac{-5}{3 \sqrt{2}}\right) \end{aligned} $

So, the D.c.'s are $\langle 2,-3,-5\rangle$.

Given points are $A(2,0,-1)$, $B(1,-2,0), C(1,2,-1)$ and $D(0,-1,-2)$.

$ \begin{aligned} & \text { So, } \begin{aligned} & \mathbf{A B}=B-A=(1,-2,0)-(2,0,-1) \\ &=(-1,-2,+1) \\ & \begin{aligned} \mathbf{A C} & =C-A=(1,2,-1)-(2,0,-1) \\ & =(-1,2,0) \end{aligned} \end{aligned} \end{aligned} $

Now, normal vector

$ \begin{aligned} & \mathbf{n}_1=\mathbf{A B} \times \mathbf{A C}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -1 & -2 & 1 \\ -1 & 2 & 0 \end{array}\right| \\ & =(0-2) \hat{\mathbf{i}}-(0+1) \hat{\mathbf{j}}+(-2-2) \hat{\mathbf{k}} \\ & =-2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-4 \hat{\mathbf{k}}=(-2,-1,-4) \\ & \text { Now, } \mathbf{A D}=D-A=(0,-1,-2)-(2,0,-1) \\ & =(-2,-1,-1) \end{aligned} $

So, normal vector $n_2=\mathbf{A C} \times \mathbf{A D}$

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

$ \begin{aligned} & =(-2-0) \hat{\mathbf{i}}-(1+0) \hat{\mathbf{j}}+(1+4) \hat{\mathbf{k}} \\ & =-2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}}=(-2,-1,5) \end{aligned} $

So, $\mathbf{n}_1 \times \mathbf{n}_2=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -2 & -1 & -4 \\ -2 & -1 & 5\end{array}\right|$

$ \begin{gathered} =(-5-4) \hat{\mathbf{i}}-(-10-8) \hat{\mathbf{j}}+(2-2) \hat{\mathbf{k}} \\ =-9 \hat{\mathbf{i}}+18 \hat{\mathbf{j}}+0 \hat{\mathbf{k}} \\ (-9,18,10) \end{gathered} $

And $\mathbf{n}_1 \cdot \mathbf{n}_2=(-2,-1,-4) \cdot(-2,-1,5)$

$ =4+1-20=-15 $

So, the tangent of the angle between the two planes

$ \begin{aligned} \tan \theta & =\frac{\left|\mathbf{n}_1 \times \mathbf{n}_2\right|}{\left|\mathbf{n}_1 \cdot \mathbf{n}_2\right|}=\frac{\sqrt{(-9)^2+(18)^2+0^2}}{|-15|} \\ & =\frac{\sqrt{81+324+0}}{|-15|}=\frac{\sqrt{405}}{15} \\ & =\frac{9 \sqrt{5}}{15}=\frac{3 \sqrt{5}}{5}=\frac{3}{\sqrt{5}} \end{aligned} $

So, the tangent of the acute angle between the two planes is $\frac{3}{\sqrt{5}}$.

$A(0,1,2), B(2,-1,3), C(1,-3,1)$

$ \begin{array}{ll} \because & \mathbf{B A}=<-2,2,-1 \geqslant \\ & \mathbf{B C}=<-1,-2,-2> \\ \because & \mathbf{B A} \cdot \mathbf{B C}=(-2)(-1)+2(-2) \\ & +(-1)(-2)=0 \end{array} $

So, $\triangle A B C$ is right angled at $B$.

$\therefore$ Orthocenter $(H)$ is at point $B=(2,-1,3)$ and circumcenter

$ (O)=\left(\frac{0+1}{2}, \frac{-3+1}{2}, \frac{1+2}{2}\right)=\left(\frac{1}{2},-1, \frac{3}{2}\right) $

Hence,

$ \begin{aligned} O H & =\sqrt{\left(2-\frac{1}{2}\right)^2+(-1+1)^2+\left(3-\frac{3}{2}\right)^2} \\ & =\sqrt{\frac{9}{4}+\frac{9}{4}}=\frac{3}{\sqrt{2}} \end{aligned} $

$\because l-2 m+n=0 \Rightarrow l=2 m-n$

Substitute in $l m+10 m n-2 n l=0$

$ \begin{aligned} & \Rightarrow(2 m-n) m+100 m n-2 n(2 m-n)=0 \\ & \Rightarrow 2 m^2+5 m n+2 n^2=0 \\ & \Rightarrow m=\frac{-5 n \pm 3 n}{4}=\frac{-n}{2},-2 n \end{aligned} $

If $m=\frac{-n}{2}, l=-2 n$

Then, direction ratios of are

$ 1: m: n=-2 n: \frac{n}{2}: n=-4:-1: 2 $

and direction cosine are

$ \begin{array}{r} (-4 k)^2+(-k)^2+(2 k)^2=1 \Rightarrow k=\frac{1}{\sqrt{21}} \\ =\left(\frac{-4}{\sqrt{21}}, \frac{-1}{\sqrt{21}}, \frac{2}{\sqrt{21}}\right) \end{array} $

If $m=-2 n$, then $l=-5 n$ and DR's are $l: m: n=-5:-2: 1$ and DC's are $(-5 k)^2+(-2 k)^2+k^2=1$

$ \Rightarrow \quad k=\frac{1}{\sqrt{30}}=\left(\frac{-5}{\sqrt{30}}, \frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}\right) $

Therefore angle between given lines are

$ \begin{aligned} & \cos \theta=\left|l_1 l_2+m_1 m_2+n_1 n_2\right| \\ & =\left\lvert\, \begin{array}{c} \frac{-4}{\sqrt{21}} \times\left(\frac{-5}{\sqrt{30}}\right)+\left(\frac{-1}{\sqrt{21}}\right) \\ \left(\frac{-2}{\sqrt{30}}\right)+\left(\frac{2}{\sqrt{21}}\right)\left(\frac{1}{\sqrt{30}}\right) \end{array}\right. \end{aligned} $

$ \begin{aligned} & =\frac{24}{\sqrt{630}}=\frac{24}{3 \sqrt{70}} \\ & =\frac{8}{\sqrt{70}} \end{aligned} $

As given, the vector from the origin to the point $(2,-1,3)$ is normal to plane.

$ \therefore \quad \mathbf{n}=<2,-1,3> $

Then, general equation of plane

$ 2 x-y+3 z=d $

$\because \quad(2,-1,3)$ lies on the plane.

$ \therefore \quad 2(2)-(-1)+3 \times 3=d \Rightarrow d=14 $

Hence, equation of the plane

$ 2 x-y+3 z-14=0 $

Given, $A=(2,-1,1), B=(2,5,1)$, $C=(0,-2,3)$

$ \begin{aligned} &\text { Using angle bisector theorem, }\\ &\frac{A B}{A C}=\frac{B D}{D C} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{\sqrt{(2-2)^2+(5-(-1))^2+(1-1)^2}}{\sqrt{(0-2)^2+(-2+1)^2+(3-1)^2}}=\frac{B D}{D C} \\ & \Rightarrow \frac{B D}{D C}=\frac{6}{3}=\frac{2}{1} \\ & \therefore B D: D C=2: 1 \end{aligned} $

Now, we find coordinates of $D$ by the help of section formula,

$ \begin{aligned} & D=\left(\begin{array}{rr} \frac{2 \times 0+1 \times 2}{2+1}, & \frac{2 \times(-2)+1 \times 5}{2+1}, \\ & \frac{2 \times 3+1 \times 1}{2+1} \end{array}\right) \\ & =\left(\frac{2}{3}, \frac{1}{3}, \frac{7}{3}\right) \end{aligned} $

Hence,

$ \begin{aligned} A D & =\sqrt{\left(2-\frac{2}{3}\right)^2+\left(-1-\frac{1}{3}\right)^2+\left(1-\frac{7}{3}\right)^2} \\ & =\sqrt{\left(\frac{4}{3}\right)^2+\left(\frac{-4}{3}\right)^2+\left(\frac{-4}{3}\right)^2} \\ & =\sqrt{\frac{48}{9}}=\sqrt{\frac{16}{3}}=\frac{4}{\sqrt{3}} \end{aligned} $

Given equation of plane $\pi$ is

$ a x+b y+11 z+d=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Now, the normal vector of $\pi$ is

$ \mathbf{n}_1=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+11 \hat{\mathbf{k}}=\langle a, b, 11\rangle $

Normal of first plane:

$ \mathbf{n}_2=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+1 \cdot \hat{\mathbf{k}}=\langle 2,-3,1\rangle $

Normal of second plane:

$ \mathbf{n}_3=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}=\langle 3,1,-1\rangle $

According to question, if plane $\pi$ is perpendicular to both then its normal vector is perpendicular to the normals of the two planes.

So, $\mathbf{n}_1 \cdot \mathbf{n}_2=0 \Rightarrow 2 a-3 b+11=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \mathbf{n}_1 \cdot \mathbf{n}_3=0 \Rightarrow 3 a+b-11=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

On solving Eqs. (ii) and (iii) we get,

$ \begin{aligned} & a=2 \text { and } b=11-3 a=11-6=5 \\ \therefore & < a, b, 11 >=< 2,5,11 > \end{aligned} $

$\therefore$ Equation of plane will become,

$ 2 x+5 y+11 z+d=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

Now, distance from origin $(0,0,0)$ to the plane (iv) is

$ \begin{aligned} & \frac{|d|}{\sqrt{2^2+5^2+11^2}}=\sqrt{6} \\ \Rightarrow \quad & |d|=\sqrt{6}+\sqrt{150}=30 \\ \Rightarrow \quad & d= \pm 30 \end{aligned} $

(given)

As question demand is all intercepts (i.e. $x, y \times z$ intercepts) are positive.

So, $d$ must be negative because $x$-intercept $=\frac{-d}{2}>0 \Rightarrow d<0 y$-intercept $=-\frac{d}{5}>0 \Rightarrow d<0$

$z$-intercept $=\frac{-d}{11}>0 \Rightarrow d<0$

Hence, $d=-30=-3 \times 2 \times 5=-3 a b$

$ \begin{aligned} & \text {Point }(-\hat{\mathbf{i}}+p \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \Rightarrow \text { Point } \\ & \text { (-1, p, -3) } \\ & \text { Plane r } \cdot(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})=7 \\ & \Rightarrow 2 x-3 y+6 z-7=0 \end{aligned} $

$ \begin{aligned} P Q & =\frac{|-2-3 p-18-7|}{\sqrt{2^2+(-3)^2+6^2}} \\ 6 & =\frac{|-3 p-27|}{\sqrt{49}} \\ 42 & =|-3 p-27| \\ \Rightarrow & (-3 p-27)= \pm 42 \\ \Rightarrow & -3 p-27=42 \\ \Rightarrow & -3 p=69 \\ \Rightarrow \quad & p=-23 \\ \Rightarrow & -3 p-27=-42 \\ \Rightarrow & -3 p=-15 \\ & p=5 \end{aligned} $

Let $P$ divide $A B$ in the ratio $m: n$.

$ \begin{aligned} & 0=\frac{2 n-m}{m+n},-7=\frac{-5 n-8 m}{m+n}, 1=\frac{3 n}{m+n} \\ & \Rightarrow 2 n-m=0 \Rightarrow m=2 n \end{aligned} $

$\Rightarrow P$ divides $A B$ in the ratio $2: 1$

Since $Q$ is the hormonic conjugate of $P$ it divides $A B$ in the ratio $-2: 1$

Using the section formula with the ratio -2:1

$ \begin{aligned} & \alpha=\frac{2-2 \times(-1)}{1-2}=-4 \\ & \beta=\frac{-5-2(-8)}{1-2}=-11 \\ & \gamma=\frac{3-2(0)}{1-2}=\frac{3}{-1}=-3 \end{aligned} $

Thus, $Q=(-4,-11,-3)$

$ \Rightarrow \quad \alpha-\beta+\gamma=-4+11-3=4 $

Let $B$ divide $A C$ in the ratio $\lambda{: 1}$ The coordinates of $B$ are given by

$ \begin{aligned} x_B & =\frac{\lambda\left(x_1+k l\right)+1 x_1}{\lambda+1} \\ y_B & =\frac{\lambda\left(y_1+k m\right)+1 y_1}{\lambda+1} \\ Z_B & =\frac{\lambda\left(z_1+k n\right)+1 z_1}{\lambda+1} \\ \Rightarrow x_1+4 k l & =\frac{\lambda\left(x_1+k l\right)+x_1}{\lambda+1} \\ \Rightarrow y_1+4 k m & =\frac{\lambda\left(y_1+k m\right)+y_1}{\lambda+1} \\ \Rightarrow z_1+4 k n & =\frac{\lambda\left(z_1+k n\right)+z_1}{\lambda+1} \end{aligned} $

$ \begin{aligned} & \text { Now, }\left(x_1+4 k l\right)(\lambda+1)=\lambda\left(x_1+k l\right)+x_1 \\ & \\ \Rightarrow& x_1 \lambda+x_1+4 k l \lambda+4 k l=x_1 \lambda+k l \lambda+x_1 \\ & \Rightarrow \quad 4 k l \lambda+4 k l=k l \lambda \\ & \Rightarrow \quad \lambda=\frac{4 k l}{-3 k l}=\frac{4}{-3} \Rightarrow \text { Ratio }=4:-3 \end{aligned} $

$(2,3,1)(1,3,2)$

$ \begin{aligned} \mathbf{n} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{array}\right| \\ & =\hat{\mathbf{i}}(6-3)-\hat{\mathbf{j}}(4-1)+\hat{\mathbf{k}}(6-3) \\ & =3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $

Since, lines of intersection of the given planes makes an angle $\alpha$ with positive $X$-axis.

$ \therefore \cos \alpha=\frac{3}{\sqrt{3^2+(-3)^2+3^2}}=\frac{3}{3 \sqrt{3}}=\frac{1}{\sqrt{3}} $

Equation of line,

$ r=(\hat{\mathbf{i}}-2 \hat{\mathbf{j}})+\lambda(2 \hat{\mathbf{i}}+\hat{\mathbf{k}}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Plane passes through $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$ and is parallel to the vectors $\mathbf{v}_1=2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$,

$ \mathbf{v}_2=\hat{\mathbf{i}}+2 \hat{\mathbf{k}} $

Any point on the plane can be written as

$ \mathbf{r}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+x(2 \hat{\mathbf{j}}-\hat{\mathbf{k}})+v(\hat{\mathbf{i}}+2 \hat{\mathbf{k}}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii),

$ \begin{aligned} & 1+2 \lambda=1+V \Rightarrow 2 \lambda=V \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & x=-2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ & \lambda=-x+2 V \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\end{aligned} $

We get $\lambda=-\frac{2}{3}$

$ \begin{aligned} \mathbf{r} & =(\hat{\mathbf{i}}-2 \hat{\mathbf{j}})-\frac{2}{3}(2 \hat{\mathbf{i}}+\hat{\mathbf{k}}) \\ & =\frac{-1}{3}(\hat{\mathbf{i}}+6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{aligned} $

$ \begin{aligned} & |\mathbf{O D}|=\sqrt{0^2+0^2+a^2}=a \\ & |\mathbf{O C}|=\sqrt{a^2+a^2+a^2}=a \sqrt{3} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad \cos \alpha & =\frac{a^2}{a \sqrt{3} \cdot a}=\frac{1}{\sqrt{3}} \\ \Rightarrow \quad \alpha & =\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ |O E| & =a \sqrt{2} \\ \cos \theta & =\frac{2 a^2}{\sqrt{2} \cdot \sqrt{3} a^2} \Rightarrow \theta=\cos ^{-1} \sqrt{\frac{2}{3}} \end{aligned} $

Plane passing through $A(1,-2,3)$ and $B(6,4,5)$

Normal vector of the given plane $\pi$

$ \mathbf{n}_1=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} $

Normal vector of plane

$ \begin{aligned} \mathbf{n} & =\mathbf{A B} \times \mathbf{n}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 5 & 6 & 2 \\ 3 & -1 & 1 \end{array}\right| \\ & =8 \hat{\mathbf{i}}+\hat{\mathbf{j}}-23 \hat{\mathbf{k}} \end{aligned} $

Equation of plane $\pi$

$ \begin{aligned} & 8(x-1)+1(y+2)-23(z-3)=0 \\ & \Rightarrow 8 x+y-23 z+63=0 \end{aligned} $

Perpendicular distances from $(0,0,0)$ to the plane $\pi$

$ d=\frac{63}{\sqrt{64+1+529}}=\frac{63}{\sqrt{594}} $

Let $P$ and $Q$ be the points on the given lines where they intersect.

$ \begin{aligned} & P=(1+t,-6+2 t, 2+t) \text { and } \\ & Q=(2 s, 4+s, 1+2 s) \end{aligned} $

Now, $P$ and $Q$ are identical

$ \begin{aligned} \therefore \quad t+1 & =2 s \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ -6+2 t & =4+s \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ 2+t & =1+2 s \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

On solving Eqs. (i) and (ii), we get

$ s=4 \text { and } t=7 $

Which satisfy the Eq. (iii)

$\therefore$ Point of intersection $(8,8,9)$

$ =8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+9 \hat{\mathbf{k}} $

Let four points $A(6,2,4), B(1,3,5)$, $C(1,-2,3)$ and $D(6, k, 2)$

$ \begin{aligned} & \mathbf{A B}=-5 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \mathbf{A C}=-5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & \mathbf{A D}=(k-2) \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \end{aligned} $

Given, four points $A B C$ and $D$ are coplanar

$ \begin{aligned} & \therefore[\mathbf{A B}, \mathbf{A C}, \mathbf{A D}]=0 \\ & \Rightarrow\left|\begin{array}{ccc} -5 & 1 & 1 \\ -5 & -4 & -1 \\ 0 & k-2 & -2 \end{array}\right|=0 \\ & \Rightarrow-5(8+k-2)-1(10)+1(-5 k+10)=0 \\ & \Rightarrow-5(6+k)-10-5 k+10=0 \\ & \Rightarrow-30-5 k-10-5 k+10=0 \\ & \Rightarrow-10 k-30=0 \\ & \therefore k=-3 \end{aligned} $

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} $

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 $