Three Dimensional Geometry

Equation of line $L$ is

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

Any point on this line can be

$ \begin{aligned} & (k,-k+3, k+2) \\ & \therefore \text { Let } A=(k,-k+3, k+2), B \equiv(0,3,2) \end{aligned} $

Now, $A B=3$

$ \Rightarrow \quad k^2+k^2+k^2=9 \Rightarrow k= \pm \sqrt{3} $

Normal to the plane is

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

DR's of normal are $\langle 1,1,-1\rangle$

Equation of normal is

$ \frac{x-\sqrt{3}}{1}=\frac{y+\sqrt{3}-3}{1}=\frac{z-\sqrt{3}-2}{-1} $

which is same as option (a).

$\pi_1$ is $\mathbf{r} \cdot(a \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})=2 a-5$

$ \begin{aligned} & \pi_2 \text { is } \mathbf{r} \cdot(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})=-4 \\ & \cos \theta=-\sqrt{\frac{3}{7}}=\frac{(a \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{a^2+4+9} \sqrt{1+4+1}} \\ & \Rightarrow \quad-\sqrt{\frac{3}{7}} \times \sqrt{6}=\frac{(a-4-3)}{\sqrt{a^2+13}} \\ & \Rightarrow \quad \frac{(a-7)^2}{a^2+13}=\frac{18}{7} \\ & \Rightarrow \quad 7 a^2-98 a+343=18 a^2+234 \\ & \Rightarrow \quad 11 a^2+98 a-109=0 \\ & \Rightarrow \quad(11 a+109)(a-1)=0 \\ & \Rightarrow \quad a=1 \text { or } a=-109 / 11 \end{aligned} $

$\therefore \quad a=1$ is the integral solution.

$A(1,3,5), B(2,4,6), C(4,5, k)$

$ \begin{aligned} \Rightarrow \quad \mathbf{A B}=(1,1,1) ; \quad \mathbf{B C}=(2,1, k-6) \\ \quad \mathbf{A C}=(3,2, k-5) \end{aligned} $

According to the question, $\mathbf{A B} \cdot \mathbf{B C}=\mathbf{0}$

$ \begin{aligned} & \Rightarrow \quad 2+1+k-6=0 \Rightarrow k=3 \\ & \operatorname{AB} \cdot \mathbf{A C}=0 \\ & \Rightarrow \quad 3+2+k-5=0 \\ & \Rightarrow \quad k=0 \end{aligned} $

$\mathbf{B C} \cdot \mathbf{A C}=\mathbf{0}$

$ \begin{array}{ll} \Rightarrow & 6+2+(k-6)(k-5)=0 \\ \Rightarrow & k^2-11 k+38=0 \\ & D<0 \end{array} $

⇒ No real values of $k$ Hence, number of possible values of $k$ is 2 .

$ \text { } \begin{aligned} \mathbf{A B} & =\hat{\mathbf{i}}+4 \hat{\mathbf{k}} \\ \mathbf{C D} & =7 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ |\mathbf{A B}| & =\sqrt{1^2+4^2}=\sqrt{17} \\ |\mathbf{C D}| & =\sqrt{49+1+1}=\sqrt{51} \\ \therefore \quad \cos \theta & =\frac{\mathbf{A B} \cdot \mathbf{C D}}{|\mathbf{A B}||\mathbf{C D}|}=\frac{7-4}{\sqrt{17} \sqrt{51}}=\frac{3}{17 \sqrt{3}} \end{aligned} $

Vector perpendicular to the plane

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

∴ DR's of normal are $\langle 1,3,-5)\rangle$.

Equation of plane

$ (x-2) \cdot 1+(y-1) \cdot 3+(z-k)(-5)=0 $

∵ It passes through the point $(3,-1,4)$.

$ \begin{array}{lrl} \therefore & (3-2)+3(-1-1)-5(4-k) & =0 \\ \Rightarrow & 1-6-20+5 k & =0 \\ \Rightarrow & & k =5 \end{array} $

Equation of line $L: \mathbf{r}=\mathbf{a}+\lambda \mathbf{b}$

$ \mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+\lambda(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}) $

$P$ be any point on line $L$.

$P(1+2 \lambda, 2-3 \lambda,-2-6 \lambda)$ and $A(1,2,-2)$

Given, $A P=21$

$ \begin{aligned} & \sqrt{(2 \lambda)^2+(-3 \lambda)^2+(-6 \lambda)^2}=21 \\ & \Rightarrow|\lambda|(\sqrt{4+9+36})=21 \\ & \Rightarrow|\lambda|=3 \Rightarrow \lambda= \pm 3 \end{aligned} $

At $\lambda=3, P(7,-7,-20)$

Position vector of $P: 7 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-20 \hat{\mathbf{k}}$

At $\lambda=-3$ and $P(-5,11,16)$

Position vector of $P:-5 \hat{\mathbf{i}}+11 \hat{\mathbf{j}}+16 \hat{\mathbf{k}}$

Given, point on plane $(1,-2,3)$

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

Perpendicular vector perpendicular to plane

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

Equation of required plane, $\mathbf{r} \cdot \mathbf{n}=\mathbf{a} \cdot \mathbf{n}$

$ \begin{aligned} & (x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \\ = & (\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \\ \Rightarrow \quad & x+2 y-3 z=-1-4-9=-14 \\ \Rightarrow \quad & x-2 y+3 z=14 \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & A(1,2,3), B(-1,4,6), C(0,-6,4), D(1,1,1) \\ & G\left(\frac{1-1+0+1}{4}, \frac{2+4-6+1}{4}\right. , \left.\frac{3+6+4+1}{4}\right) \equiv G\left(\frac{1}{4}, \frac{1}{4}, \frac{14}{4}\right) \\ & G_1\left(\frac{-1+0+1}{3}, \frac{4-6+1}{3}, \frac{6+4+1}{3}\right) \\ & \equiv G_1\left(0, \frac{-1}{3}, \frac{11}{3}\right) \\ & A G_1=\sqrt{1^2+\left(\frac{7}{3}\right)^2+\left(\frac{2}{3}\right)^2} \\ & \quad=\frac{1}{3} \sqrt{9+49+4}=\frac{\sqrt{62}}{3} \\ & A G=\sqrt{\left(\frac{3}{4}\right)^2+\left(\frac{7}{4}\right)^2+\left(\frac{2}{4}\right)^2}=\frac{\sqrt{62}}{4} \\ & \frac{A G_1}{A G}=\frac{\sqrt{62} / 3}{\sqrt{62} / 4}=\frac{4}{3} \end{aligned} \end{aligned} $

Here, $P_1: x-y+z+2=0$

$ \begin{aligned} & n_1: 1,-1,1 \\ & P_2: 2 x+y-2 z+5=0 \\ & n_2: 2,1,-2 \end{aligned} $

vector along line of intersection of $P_1$ and $P_2=\mathbf{n}_1 \times \mathbf{n}_2$

$ \begin{aligned} & \mathrm{n}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 2 & 1 & -2 \end{array}\right|=\mathbf{i}+4 \mathbf{j}+3 \mathbf{k} \\ & |\mathrm{n}|=\sqrt{1+16+9}=\sqrt{26} \\ & \text { DC's are } \frac{1}{\sqrt{26}}, \frac{4}{\sqrt{26}}, \frac{3}{\sqrt{26}} \end{aligned} $

$ \text { Dr's of normal to plane } $

$ \begin{aligned} & : 1-(-1), 2-3,3-(-2) \\ & :(2,-1,5) \equiv(a, b, c) \end{aligned} $

Plane passes through

$ (-1,3,-2) \equiv\left(x_1, y_1, z_1\right) $

Equation of plane

$ \begin{array}{r} a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0 \\ 2(x+1)-(y-3)+5(z+2)=0 \\ 2 x-y+5 z+15=0 \end{array} $

Perpendicular distance from origin to the plane

$ \left|\frac{0-0+0+15}{\sqrt{4+1+25}}\right|=\frac{15}{\sqrt{30}}=\sqrt{\frac{15}{2}} $

Given, points $A(1,2,-1)$ and $B(-1,0,1)$ and $P$ divides the line segment externally in the ratio $1: 2$.

For external division, coordinates are

$\begin{aligned} & {\left[\frac{m_1 x_2-m_2 x_1}{m_1-m_2}, \frac{m_1 y_2-m_2 y_1}{m_1-m_2}, \frac{m_1 z_2-m_2 z_1}{m_1-m_2}\right]} \\ & \quad=\left[\frac{1(-1)-2(1)}{1-2}, \frac{1(0)-2(2)}{1-2}, \frac{1(1)-2(-1)}{1-2}\right] \\ & \quad=\left[\frac{-1-2}{-1}, \frac{-4}{-1}, \frac{3}{-1}\right] \\ & \quad=[3,4,-3] \end{aligned}$

$\begin{aligned} \text{So,}\quad PQ & =\sqrt{(3-1)^2+(4-3)^2+(-3+1)^2} \\ & =\sqrt{(2)^2+(1)^2+(-2)^2} \\ & =\sqrt{4+1+4}=3 \text { units } \end{aligned} $

Direction cosine of $a \hat{i}+b \hat{j}+c \hat{k}$ can be $\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}$

According to question,

$\sqrt{a^2+b^2+c^2}=\sqrt{83}, b=5$

$\begin{array}{lr} \Rightarrow & \quad a^2+(5)^2+c^2=83 \\ \Rightarrow & a^2+c^2=83-25 \\ \Rightarrow & a^2+c^2=58 \quad \text{... (i)} \end{array}$

and $c-a=4\quad \text{[given]}$

$\begin{aligned} &\begin{aligned} & (c-a)^2+2 a c=a^2+c^2 \\ & \Rightarrow(4)^2+2 a c=58 \quad\text { [from Eq. (i), } c^2+a^2=58 \text { ] }\\ & \Rightarrow 2 a c=58-16 \\ & a c=\frac{42}{2}=21 \end{aligned}\\ \end{aligned}$

The equation of plane passes through the point $(1,0,1)$ is

$a(x-1)+b(y-0)+c(z-1)=0\quad \text{... (i)}$

If this plane is perpendicular to the planes $2 x+3 y-z=2 \text { and } x-y+2 z=1$

Then, $2 a+3 b-c=2$ and $a-b+2 c=1$

$\begin{aligned} & \Rightarrow \frac{a}{6-1}=\frac{b}{-1-4}=\frac{c}{2(-1)-3} \\ & \Rightarrow \frac{a}{5}=\frac{b}{-3}=\frac{c}{-5}=k \end{aligned}$

Let $a=5 k, b=-3 k, c=-5 k$

Putting the values of $a, b$ and $c$ in Eq. (i), we get the required equation of plane

$\begin{aligned} & 5 k(x-1)+(-3 k)(y)+(-5 k)(z-1)=0 \\ \Rightarrow & 5 x-5-3 y-5 z+5=0 \\ \Rightarrow & 5 x-3 y-5 z=0 \end{aligned}$

Now, plane passing through $(11,7,5)$ and parallel to the plane $\pi(5 x-3 y-5 z)$.

Equation of a plane parallel to

$5 x-3 y-5 z=k \quad \text{... (i)}$

and passing through $(11,7,5)$

$\begin{aligned} & & 5(11)-3(7)-5(5) & =k \\ \Rightarrow & & 55-21-25 & =k \\ \Rightarrow & & k & =9 \end{aligned}$

So, equation of plane

$\Rightarrow 5 x-3 y-5 z=9$

Comparing with $a x+b y-z+d=0$ (given)

So, $a=5, b=-3, c=5, d=9$

$\begin{aligned} & \frac{a}{b}+\frac{b}{d}=\frac{5}{-3}+\frac{-3}{9} \\ \Rightarrow & \frac{a}{b}+\frac{b}{d}=\frac{-15-3}{9}=\frac{-18}{9}=-2 \end{aligned}$

Let a, b and c are the position vectors of the vertices A, B are C, respectively.

Also $\mathbf{d}, \mathbf{e}$ and $\mathbf{f}$ be the position vectors of points $D, E$ and $F$, respectively

$\mathrm{d}=\frac{2 \mathrm{c}+3 \mathrm{~b}}{5}, \mathrm{e}=\frac{2 \mathrm{c}+\mathrm{a}}{3}, \mathrm{f}=\frac{3 \mathrm{~b}+\mathrm{a}}{4}$

Equation of line $B E$ is $\mathbf{r}=\mathbf{b}+k(\mathbf{e}-\mathbf{b})$,

$\begin{aligned} \mathbf{r} & =\mathbf{b}+k\left(\frac{2 \mathbf{c}+\mathbf{a}}{3}-\mathbf{b}\right) \\ \Rightarrow \quad \mathbf{r} & =\mathbf{b}+k\left[\frac{2 \mathbf{c}+\mathbf{a}-3 \mathbf{b}}{3}\right] \\ \Rightarrow \quad \mathbf{r} & =\frac{k}{3} \mathbf{a}+\mathbf{b}[1-k]+\frac{2 k}{3} \mathbf{c} \end{aligned}$

Equation of line $C F$ is

$\begin{aligned} \mathbf{r} & =\mathbf{c}+\lambda(\mathbf{f}-\mathbf{c}) \\ & =\mathbf{c}+\lambda\left(\frac{3 b+\mathbf{a}}{4}-\mathbf{c}\right)=\mathbf{c}+\lambda\left[\frac{3 b+\mathbf{a}-4 \mathbf{c}}{4}\right] \\ & =\frac{\lambda \mathbf{a}}{4}+\frac{3 \lambda \mathbf{b}}{4}+\mathbf{c}(1-\lambda) \end{aligned}$

$C F$ and $B E$ intersect at point $P$

$\frac{k \mathbf{a}}{3}+\mathbf{b}(1-k)+\frac{2 k}{3} \mathbf{c}=\frac{\lambda \mathbf{a}}{4}+\frac{3 \lambda \mathbf{b}}{4}+\mathbf{c}(1-\lambda)$

On comparing,

$\frac{k}{3}=\frac{\lambda}{4}$

$\Rightarrow \quad k=\frac{3 \lambda}{4}$

Also,

$\begin{array}{rlrl} & & 1-k & =\frac{3 \lambda}{4} \Rightarrow 1-k=k \\ \Rightarrow & & 2 k & =1 \Rightarrow k=\frac{1}{2} \\ \text{and} & & \frac{1}{2} & =\frac{3 \lambda}{4} \\ && \lambda & =\frac{2}{3} \end{array}$

Position vector of $P=\frac{\mathbf{a}}{6}+\frac{\mathbf{b}}{2}+\frac{\mathbf{c}}{3}=\frac{\mathbf{a}+3 \mathbf{b}+2 \mathbf{c}}{6}$

$\begin{aligned} & \mathbf{A P}=x_1 \mathbf{A B}+y_1 \mathbf{A C} \\ &\left(\frac{\mathbf{a}+3 \mathbf{b}+2 \mathbf{c}}{6}-\mathbf{a}\right)=x_1(\mathbf{b}-\mathbf{a})+y_1(\mathbf{c}-\mathbf{a}) \\ & \Rightarrow \frac{-5 \mathbf{a}+3 \mathbf{b}+2 \mathbf{c}}{6}=\mathbf{a}\left(-x_1-y_1\right)+x_1 \mathbf{b}+y_1 \mathbf{C} \\ & \therefore \quad-x_1-y_1=\frac{-5}{6} \\ & x_1+y_1=\frac{5}{6} \end{aligned}$

The equation of the plane passing through the point $A(-2,1,3)$ and perpendicular to the vector $3 \hat{i}+\hat{j}+5 \hat{k}$ is $3 x+y+5 z=d$.

Thus, $d=-6+1+15=10$

Then, $3 x+y+5 z-10=0$

Thus, $a=3, b=1, c=5, d=-10$

$\frac{a+b}{c+d}=\frac{3+1}{5-10}=-\frac{4}{5}$

Given, $Q(2,2,1)$ and $R(5,2,-2)$.

Thus, equation of line through $R$ and $Q$ is

$\begin{aligned} & \frac{x-2}{2-5}=\frac{y-2}{2-2}=\frac{z-1}{1+2} \\ \Rightarrow & \frac{x-2}{-3}=\frac{y-2}{0}=\frac{z-1}{3}=r \text { (say) } \end{aligned}$

Thus, $P$ be point on the line. So, $P=(-3 r+2), 2,(3 r+1)$

Since, $x$-coordinate is 4 . So, $-3 r+2=4$

$\Rightarrow \quad r=\frac{-2}{3}$

Thus, $y$-coordinate $=2$

and $z$-coordinate $=-1$

$\therefore y$-coordinate $=-2(z$ - coordinate of $P$)

Mid-point of $A$ and $B$ is $\left(\frac{p+3}{2}, \frac{-4+2}{2}, \frac{2-4}{2}\right) =\left(\frac{p+3}{2},-1,-1\right)$

Let $P$ be mid-point of $A$ and $B$.

DR'S of $C P=\frac{p+3}{2}-5,-1-q,-1-1$

$=\frac{p-7}{2},-1-q,-2$

$\begin{aligned} & \text { Here, } \frac{p-7}{2}=2,-1-q=3, c=-2 \\ & \Rightarrow \quad p=11, q=-4, c=-2 \\ & \Rightarrow \quad c(p+7 q)=-2[11-28]=-2(-17)=34 \end{aligned}$

Given, the plane normal to the vector

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

Let equation of plane

$x+2 y+2 z+d=0$

Distance of a plane from the origin is $1 / 3$.

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

$\begin{aligned} & \Rightarrow \frac{1}{3}=\left|\frac{0+2 \cdot 0+2 \cdot 0+d}{\sqrt{1+2^2+2^2}}\right| \\ & \Rightarrow 1 / 3=d / 3 \Rightarrow d=1 \end{aligned}$

Thus, equation of the plane is $x+2 y+2 z+1=0$

Comparing given equation of the plane

$\begin{aligned} & x+p y+q z+r=0 \\ & \therefore \quad p=2, q=2, r=1 \\ & \therefore \quad \sqrt{p^2+q^2+r^2}=\sqrt{2^2+2^2+1^2} \\ & \Rightarrow \quad \sqrt{9}=3 \\ \end{aligned}$

$\mathbf{r}=2 \mathbf{b}+t(6 \mathbf{c}-\mathbf{a})$

and $\mathbf{r}=\mathbf{a}+s(\mathbf{b}-3 c)$

Put $t=-1$ and $s=24$ and check

If $t=-1 \Rightarrow \mathbf{r}=2 \mathrm{~b}-1(6 \mathrm{c}-\mathrm{a})=2 \mathrm{~b}+\mathrm{a}-6 \mathrm{c}$

If $s=2 \Rightarrow \mathbf{r}=\mathbf{a}+2(\mathbf{b}-3 \mathrm{c})=\mathbf{a}+2 \mathbf{b}-6 \mathbf{c}$

Therefore, we can say that $\mathbf{a}+2 b-6 c$ lies on both the lines, i.e. it is the intersecting point.

$\begin{aligned} x & =\frac{m x_2+n x_1}{m+n} \\ \Rightarrow \quad a & =\frac{5 m+n}{m+n} \\ \Rightarrow \quad a & =\frac{5\left(\frac{m}{n}\right)+1}{\frac{m}{n}+1} \quad \text{.... (i)} \end{aligned}$

$\begin{aligned} & y \text {-coordinate }: 8=\frac{2 m+4 n}{m+n} \\ & \Rightarrow \quad 8 m+8 n=2 m+4 n \\ & \Rightarrow \quad 6 m=-4 n \\ & \Rightarrow \quad 3 m=-2 n \quad \text{... (ii)}\\ \end{aligned}$

$z \text { - coordinate }:-2=\frac{10 m+6 n}{m+n}$

$\begin{aligned} \Rightarrow & -2 m-2 n =10 m+6 n \\ \Rightarrow & -8 n =12 m \Rightarrow-2 n=3 m \\ \Rightarrow & \frac{m}{n} =-\frac{2}{3}\quad \text{... (iii)} \end{aligned}$

From Eq. (iii), put $\frac{m}{n}=\frac{-2}{3}$ into Eq. (i),

$\begin{aligned} a & =\frac{5\left(-\frac{2}{3}\right)+1}{-\frac{2}{3}+1}=\frac{-10+3}{-2+3}=-7 \\ \therefore \quad & \frac{2 m}{n}-\frac{a}{3}=2\left(-\frac{2}{3}\right)-\left(-\frac{7}{3}\right) \\ & =\frac{-4}{3}+\frac{7}{3}=\frac{-4+7}{3}=\frac{3}{3}=1 \end{aligned}$

Given, points are $(4,3,-5)$ and $(-2,1,-8)$.

Direction ratios

$\begin{aligned} &(a, b, c)=(4-(-2), 3-1,-5-(-8))=(6,2,3) \\ & \therefore \quad P(a, 3 b, 2 c) \equiv(6,3 \times 2,2 \times 3) \equiv(6,6,6) \\ & \text { For } x+y-2 z=0 \\ & \Rightarrow \quad 6+6-2 \times 6=0 \\ & \Rightarrow \quad 12-12=0 \text { (true) } \end{aligned}$

Let plane's equation be

$\begin{aligned} \quad & \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \quad \text{....(i)}\\ \Rightarrow \quad & a=\frac{5}{2} \end{aligned}$

and plane passes through $(1,1,1)$.

$\begin{array}{ll} \therefore & \frac{1}{5 / 2}+\frac{1}{b}+\frac{1}{c}=1 \\ \Rightarrow & \frac{1}{b}+\frac{1}{c}=1-\frac{2}{5} \\ \Rightarrow & \frac{1}{b}+\frac{1}{c}=\frac{3}{5} \quad \text{.... (ii)} \end{array}$

Perpendicular distance of plane from origin $=\frac{5}{7}$

$\begin{aligned} & \left|\frac{0+0+0-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\right|=\frac{5}{7} \\ & \Rightarrow \quad\left|\frac{-1}{\sqrt{\left(\frac{2}{5}\right)^2+\frac{1}{b^2}+\frac{1}{c^2}}}\right|=\frac{5}{7} \\ & \Rightarrow \quad(-1)^2\left(\frac{7}{5}\right)^2=\left(\frac{2}{5}\right)^2+\frac{1}{b^2}+\frac{1}{c^2} \\ & \Rightarrow \quad \frac{49}{25}-\frac{4}{25}=\frac{1}{b^2}+\frac{1}{c^2} \\ & \Rightarrow \quad \frac{1}{b^2}+\frac{1}{c^2}=\frac{45}{25}=\frac{9}{5} \quad \text{... (iii)}\\ \end{aligned}$

On squaring Eq. (ii), we get $ \begin{aligned} & \frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{b c}=\frac{9}{25} \\ & \Rightarrow \quad \frac{9}{5}+\frac{2}{b c}=\frac{9}{25} \quad \text{[from Eq. (iii)]}\\ & \Rightarrow \quad \frac{2}{b c}=\frac{-36}{25} \\ & \Rightarrow b c=-\frac{25}{18} \end{aligned}$

Putting $c=-\frac{25}{18} \cdot \frac{1}{b}$ into Eq. (ii), we get

$\frac{1}{b}+\frac{1}{\left(\frac{-25}{18} \cdot \frac{1}{b}\right)}=\frac{3}{5}$

$\begin{array}{lrl} \Rightarrow & \frac{1}{b}-\frac{18 b}{25} =\frac{3}{5} \\ \Rightarrow & 25-18 b^2 =15 b \\ \Rightarrow & 18 b^2+15 b-25=0 \\ \Rightarrow & 18 b^2+30 b-15 b-25=0 \end{array}$

$\begin{array}{lrl} \Rightarrow & 6 b(3 b+5)-5(3 b+5) =0 \\ \Rightarrow & (3 b+5)(6 b-5) =0 \\ \Rightarrow & b =-\frac{5}{3}, \frac{5}{6} \end{array}$

$\because y$-intercept is negative.

$b=-\frac{5}{3}$

Given plane $3x + y + 2z + 6 = 0$

and line ${{x - {1 \over 3}} \over {{{2b} \over 3}}} = {{y - 3} \over { - 1}} = {{z - 1} \over a}$

Since, plane is parallel to line, then

$3\left( {{{2b} \over 3}} \right) + (1)( - 1) + 2(a) = 0$

$2b - 1 + 2a = 0$

$ \Rightarrow a + b = {1 \over 2}$

Now, $3a + 3b = {3 \over 2}$

Given that, equation of plane parallel to vectors $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$. Also, it passes through $3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$.

Let the plane be

$a(x-3)+b(y-2)+c(z-6)=0$ ...... (i)

such that,

$\mathrm{t} a(2)+b(\mathrm{l})+c(\mathrm{l}) =0$

$\Rightarrow 2a+b+c =0$ ....... (ii)

and $a(\mathrm{l})+b(-\mathrm{l})+c(\mathrm{l}) =0$

$\Rightarrow \quad a-b+c=0$ ....... (iii)

From Eq. (ii) and (iii)

$\Rightarrow \quad \frac{a}{2}=\frac{b}{-1}=\frac{c}{-3}=\lambda \quad (\text{say})$

$\Rightarrow a=2 \lambda, b=-\lambda, c=-3 \lambda$

From Eq. (i), we get

$\begin{array}{ccrl} & 2 \lambda(x-3)-\lambda(y-2)-3 \lambda(z-6)=0 \\ \Rightarrow & 2(x-3)-(y-2)-3(z-6)=0 \\ \Rightarrow & 2 x-y-3 z=-14 \end{array}$

Given point, $A(-2,4,-5)$ and $B(1,2,3)$

$A B=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}$

DR's are $(3,-2,8)$

Then, direction cosines $=(l, m, n)$

where, $l=\frac{3}{\sqrt{3^2+(-2)^2+8^2}}=\frac{3}{\sqrt{77}}$

$\begin{array}{r} m=\frac{(-2)}{\sqrt{77}} \text { and } n=\frac{8}{\sqrt{77}} \\ \therefore \text { DC's are }\left(\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}\right) \end{array}$

Given, points $(2,3,4),(-1,-2,1)$ and $(5,8,7)$

If $\Delta=0$, then points are collinear.

$\Delta \neq 0$, then it represent vertices of triangle

$\begin{aligned} & \Delta=\left|\begin{array}{ccc} 2 & 3 & 4 \\ -1 & -2 & 1 \\ 5 & 8 & 7 \end{array}\right| \\ & =2(-14-8)-3(-7-5)+4(-8+10) \\ & =-44+36+8=0\end{aligned}$

$\Rightarrow$ Given points are collinear.

Given equation of plane is

$4x+3y+2z=2$ ....... (i)

General form is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, where $a, b$ and $c$ are intercept on coordinate axis, convert Eq. (i) in general form as

$\frac{4}{2} x+\frac{3}{2} y+\frac{2}{2} z=1$

or $\quad 2 x+\frac{3}{2} y+z=1 \Rightarrow \frac{x}{1 / 2}+\frac{y}{2 / 3}+\frac{z}{1}=1$

$\begin{aligned} & \therefore x \text {-intercept }=\frac{1}{2}=(a) \\ & y \text {-intercept }=\frac{2}{3}=(b) \\ & z \text {-intercept }=1=(c) \end{aligned}$

Then, $a+b+c=\frac{1}{2}+\frac{2}{3}+1=\frac{3+4+6}{6}=\frac{13}{6}$

Since we know that two lines are coplanar if

$\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right|=0$ .... (i)

Given lines are

$\begin{aligned} L_1: \frac{x-3}{2} & =\frac{y-2}{3}=\frac{z-1}{6} \\ \therefore \quad x_1 & =3, y_1=2, z_1=1 \\ a_1 & =2, b_1=3, c_1=\lambda \\ L_2: \frac{x-z}{3} & =\frac{y-3}{2}=\frac{z-2}{3} \\ a_2 & =3, b_2=2, c_2=3 \end{aligned}$

From Eq. (i)

$\begin{aligned} & \left|\begin{array}{rrr} 1 & -1 & -1 \\ 2 & 3 & \lambda \\ 3 & 2 & 3 \end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 & 5 & \lambda+2 \\ 3 & 5 & 6 \end{array}\right|=0 \\ & \Rightarrow \quad 30-5 \lambda-10=0 \\ & \Rightarrow \quad \lambda=4 \\ & \Rightarrow \sin ^{-1}(\sin 4)+\cos ^{-1}(\cos 4) \\ & (\pi-4)+(2 \pi-4)=(3 \pi-8) \\ & \end{aligned}$

Line passing through $(1,1,-1)$ and parallel to $(1,2,-1)$ is

$\begin{aligned} & L_1: \frac{x-1}{1}=\frac{y-1}{2}=\frac{z+1}{-1}=\lambda \\ & L_2: \frac{x-3}{-1}=\frac{y+2}{5}=\frac{z-2}{-4}=\propto \end{aligned}$

For $A:(\lambda+1,2 \lambda+1,-\lambda-1)$

$\begin{aligned} & \quad=(-\propto+3,5 \propto-2,-4 \propto+2) \\ & \Rightarrow \quad \lambda+\propto=2 \Rightarrow 2 \lambda+1=5 \propto-2 \\ & \Rightarrow \quad 4-2 \propto+1=5 \propto-2 \\ & 1=\propto \quad \text { and } \lambda=1 \\ & A:(2,3,-2) \\ & P: 2 x-y+2 z+7=0 \\ & \Rightarrow 2(\lambda+1)-(2 \lambda+1)+(-2 \lambda-2)+7=0 \\ & \Rightarrow \quad-2 \lambda+6=0 \Rightarrow \lambda=3 \\ & B:(4,7,-4)\\& AB^2=2^2+4^2+2^2=24=2 \sqrt{6} \end{aligned}$

Centroid = (2, 2, 2)

$\because$ Triangle is equilateral

(2, 2, 2) = H = G = S = I

$\therefore$ H + G + S + I = (8, 8, 8)

AB = $\lambda$CD

$\Rightarrow (2,2,3)=\lambda(2,2,k-3)$

$\Rightarrow k-3=3\Rightarrow k=6$

P = Angle bisector

$ = {1 \over 2}\left[ {\left( {{2 \over 3} \pm {5 \over {13}}} \right),\left( {{2 \over 3} \pm {{12} \over {13}}} \right),\left( {{1 \over 3} \pm 0} \right)} \right] = {1 \over 2}\left( {{{41} \over {39}},{{62} \over {39}},{1 \over 3}} \right)$

Parallel to this vector will be (41, 62, 13).

Equation of plane containing line of intersection of the given planes is $(x-2y+z+2)+\lambda(3x-y-z+1)=0$

This plane also passes through (1, 1, 1).

$\therefore (1-2+1+2)+\lambda(3-1-1+1)=0$

$\Rightarrow \lambda=-1$

$\therefore$ Required plane $2x+y-2z-1=0$

$x$-intercept of these plane is (x, 0, 0)

$\Rightarrow 2x-1=0$

$x=\frac{1}{2}$

Given that line L$_1$ is passing through $(2 \hat{\mathbf{i}}-\hat{\mathbf{k}})$ and parallel to $(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})$

Let $\mathbf{a}_1=2 \hat{\mathbf{i}}-\hat{\mathbf{k}}$ and $\mathbf{b}_1=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$

equation line $L_1: \mathbf{r}=\mathbf{a}_1+\lambda \mathbf{b}_1$ ..... (i)

Also given line $L_2$ is passing through $(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})$ and parallel to $(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$

Let $\mathbf{a}_2=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ and $\hat{\mathbf{b}}_2=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$

Equation of line $L_2: \mathbf{r}=\mathbf{a}_2+\mathbf{b}_2$ ..... (ii)

$\therefore$ Shortest distance between $L_1$ and $L_2$

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

Consider $\mathbf{b}_1 \times \mathbf{b}_2=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & -1 & 2 \\ 1 & -2 & 1\end{array}\right|=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}-5 \hat{\mathbf{k}}$

$\left|\mathbf{b}_1 \times \mathbf{b}_2\right|=\sqrt{3^2+(-1)^2+(-5)^2}=\sqrt{35} $

and $\quad \mathbf{a}_2-\mathbf{a}_1=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})-(2 \hat{\mathbf{i}}-\hat{\mathbf{k}})$

$=\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$

$\begin{aligned} \left(\mathbf{b}_1 \times \mathbf{b}_2\right) \cdot\left(\mathbf{a}_2-\mathbf{a}_1\right) & =(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}-5 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\ & =-1+10=9 \end{aligned}$

Now from Eq. (iii)

$\begin{aligned} d & =\left|\frac{9}{\sqrt{35}}\right| \\ \Rightarrow \quad d & =\frac{9}{\sqrt{35}} \end{aligned}$

In parallelogram, diagonals bisect each other

Let fourth vertex (x, y, z)

Mid-point of $A C=$ mid-point of $B D$

$\begin{aligned} & \left(\frac{2+4}{2}, \frac{4+5}{2}, \frac{-1+1}{2}\right) =\left(\frac{x+3}{2}, \frac{y+6}{2}, \frac{z-1}{2}\right) \\ \Rightarrow \quad & \left(3,\frac{9}{2},0\right) =\left(\frac{x+3}{2}, \frac{y+6}{2}, \frac{z-1}{2}\right) \\ \Rightarrow \quad & \frac{x+3}{2} =3, \frac{y+6}{2}=\frac{9}{2}, \frac{z-1}{2}=0 \\ \Rightarrow \quad & x =3, y=3, z=1 \end{aligned}$

Fourth vertex $(3,3,1)$

$\mathbf{A B}=\mathbf{a}=\mathbf{B}-\mathbf{A}$

$ = (5\widehat i - 6\widehat k) - ( - \widehat i + 2\widehat j - 3\widehat k)$

$ = 6\widehat i - 2\widehat j - 3\widehat k$

$AC = c = C - A$

$ = (4\widehat j - \widehat k) - ( - \widehat i + 2\widehat j - 3\widehat k)$

$ = \widehat i + 2\widehat j + 2\widehat k$

Internal angle bisector $=\hat{\mathbf{c}}+\hat{\mathbf{a}}$

$\begin{aligned} & =\frac{\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{\sqrt{1+4+4}}+\frac{6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}}{\sqrt{36+4+9}} \\ & =\frac{\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{3}+\frac{6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}}{7} \\ & =\frac{7 \hat{\mathbf{i}}+18 \hat{\mathbf{i}}}{21}+\frac{14 \hat{\mathbf{j}}-6 \hat{\mathbf{j}}}{21}+\frac{14 \hat{\mathbf{k}}-9 \hat{\mathbf{k}}}{21} \\ & =\frac{25 \hat{\mathbf{i}}}{21}+\frac{8 \hat{\mathbf{j}}}{21}+\frac{5 \hat{\mathbf{k}}}{21} \end{aligned}$

DR's of internal bisector of

$\angle B A C=\frac{25}{21}, \frac{8}{21}, \frac{5}{21}$

Direction Cosines $=\frac{25}{\sqrt{714}}, \frac{8}{\sqrt{714}}, \frac{5}{\sqrt{714}}$

Let $\overline{A B}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$

$a, b$ and $c$ are projections of $\overline{A B}$ on $x, y$ and $z$ respectively

Projection of $\overline{A B}$ on $x y$-plane

$=\sqrt{a^2+b^2}=\sqrt{15}$

Projection of $\overline{A B}$ on $y z$-plane

$=\sqrt{b^2+c^2}=\sqrt{46}$

Projection of $\overline{A B}$ on $z x$-plane

$=\sqrt{c^2+a^2}=7$

$\therefore \quad a^2+b^2=15, b^2+c^2=46, c^2+a^2=49$

On solving $a=3, b=\sqrt{6}, c=\sqrt{40}$

Projection of $\overline{A B}$ on $Y$-axis $=b=\sqrt{6}$

* no option matches *

Let Equation of plane passing through $(2,1,3)$ is

$l(x-2)+m(y-1)+n(z-3)=0$ .... (i)

This plane is perpendicular to

$\begin{aligned} x-2 y+2 z+3 & =0 \\ \text {and} \quad 3 x-2 y+4 z-4 & =0 \\ \Rightarrow \quad l-2 m+2 n & =0 \\ \text {and} \quad 3 l-2 m+4 n & =0 \\ \Rightarrow \quad \frac{l}{-8+4} & =\frac{m}{6-4}=\frac{n}{-2+6} \\ \Rightarrow \quad \frac{l}{-4} & =\frac{m}{2}=\frac{n}{4} \Rightarrow \frac{l}{-2}=\frac{m}{1}=\frac{n}{2} \end{aligned}$

From Eq. (i),

$\begin{array}{lr} \therefore & -2(x-2)+(y-1)-2(z-3)=0 \\ \Rightarrow & -2 x+4+y-1+2 z-6=0 \\ \Rightarrow & -2 x+y+2 z-3=0 \\ \Rightarrow & 2 x-y-2 z+3=0 \end{array}$

Let the line joining the points $A(2,4,5)$ and $B(3,5,-4)$ is divided by $Y Z$-plane at $(0, y, z)$ in the ratio $m: n$. Then, by section formula $\frac{2 n+3 m}{m+n}=0$

$\begin{aligned} & \Rightarrow \quad \frac{m}{n}=-\frac{2}{3} \\ & \Rightarrow \quad m: n=2: 3 \end{aligned}$

(externally)

Let the line makes angle $\alpha, \beta, \gamma$ with the coordinate axes, then

$\begin{aligned} & \quad \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\ & \alpha=\beta=\gamma \quad \text { (given) }\\ & \Rightarrow \cos ^2 \alpha+\cos ^2 \alpha+\cos ^2 \alpha=1 \\ & \begin{aligned} \text { or } \end{aligned} \\ & 3 \cos ^2 \alpha=1 \text { or } \cos \alpha= \pm \frac{1}{\sqrt{3}} \\ & \text { Direction cosines are }\left( \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}\right) \end{aligned}$

Let $P(x, y, z)$ and $O(0,0,0)$

DR's of $O P=(x, y, z)$

But it is given that DR's of $O P=(1,-2,-2)$

$\because x=1, y=-2, z=-2$

$\therefore$ Coordinate of $P=(1,-2,-2)$

Let each edge of cube be a, then coordinates of the vertices of cube are

O(0, 0, 0), A(a, 0, 0), B(0, a, 0), C(0, 0, a), N(a, a, 0), P(a, a, a), L(0, a, a), M(a, 0, a)

Direction ratios of the diagonals OP, AL, BM and CN are (a, a, a), ($-$a, a, a), (a, $-$a, a) and (a, a, $-$a).

Let $\theta$ be the acute angle between diagonals OP and AL.

$\therefore$ $\cos \theta = {{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}} \over {\sqrt {a_1^2 + b_1^2 + c_1^2} \sqrt {a_2^2 + b_2^2 + c_2^2} }}$

$ = {{a \times ( - a) + a \times a + a \times a} \over {\sqrt {{a^2} + {a^2} + {a^2}} \sqrt {{{( - a)}^2} + {a^2} + {a^2}} }}$

$ = {{ - {a^2} + {a^2} + {a^2}} \over {\sqrt {3{a^2}} \sqrt {3{a^2}} }} = {{{a^2}} \over {a\sqrt 3 \times a\sqrt 3 }}$

$ \Rightarrow \cos \theta = {1 \over 3} \Rightarrow \theta = {\cos ^{ - 1}}\left( {{1 \over 3}} \right)$

Given lines are

${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$ and ${L_2}:{{x - 2} \over 1} = {{y + 2} \over 2} = {{z - 3} \over 3}$

Now, convert into vector form

${L_1}:( - \widehat i - 2\widehat j - \widehat k) + \lambda (3\widehat i + \widehat j + 2\widehat k)$

${L_2}:(2\widehat i + 2\widehat j + 3\widehat k) + \mu (\widehat i + 2\widehat j + 3\widehat k)$

Line L₁ comparing with ${a_1} + \lambda {b_1}$

and L₂ comparing with ${a_2} + \mu {b_2}$, then we have ${b_1} = 3\widehat i + \widehat j + 2\widehat k$ and ${b_2} = \widehat i + 2\widehat j + 3\widehat k$

Perpendicular to both b₁ and b₂ = ${b_1} \times {b_2}$

$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & 1 & 2 \cr 1 & 2 & 3 \cr } } \right|$

$ = \widehat i(3 - 4) - \widehat j(9 - 2) + \widehat k(6 - 1) = - \widehat i - 7\widehat j + 5\widehat k$

$\therefore$ Required unit vector $ = {{ - \widehat i - 7\widehat j + 5\widehat k} \over {\sqrt {{{( - 1)}^2} + {{( - 7)}^2} + {{(5)}^2}} }}$

$ = {{ - \widehat i - 7\widehat j + 5\widehat k} \over {\sqrt {75} }} = {{ - \widehat i - 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}$

The given line is r = a + tb, where $a = 2\widehat i - 2\widehat j + 3\widehat k$, $b = \widehat i - \widehat j + 4\widehat k$ and given plane is r . n = d, where $n = \widehat i + 5\widehat j + \widehat k$, d = 5.

Since, $b\,.\,n = 1 - 5 + 4 = 0$

$\therefore$ Given line is parallel to given plane.

$\therefore$ The distance between the line and the plane is equal to length of the perpendicular from the point a = ${2\widehat i - 2\widehat j + 3\widehat k}$ on the line to the given plane.

$\therefore$ Required distance $ = \left| {{{(2\widehat i - 2\widehat j + 3\widehat k)\,.\,(\widehat i + 5\widehat j + \widehat k) - 5} \over {\sqrt {1 + 25 + 1} }}} \right|$

$ = \left| {{{2 - 10 + 3 - 5} \over {\sqrt {27} }}} \right| = {{10} \over {3\sqrt 3 }}$

Given that, points $A(4,7,8), B(2,3,4)$ and $C(2,5,7)$

Length

$ \begin{aligned} A B & =\sqrt{\left(x_A-x_B\right)^2+\left(y_A-y_B\right)^2+\left(z_A-z_B\right)^2} \\ & =\sqrt{(4-2)^2+(7-3)^2+(8-4)^2} \end{aligned} $

$ \begin{aligned} & =\sqrt{(2)^2+(4)^2+(4)^2}=\sqrt{4+16+16} \\ A B & =6 \end{aligned} $

Length

$ \begin{aligned} A C & =\sqrt{\left(x_A-x_C\right)^2+\left(y_A-y_C\right)^2+\left(z_A-z_C\right)^2} \\ & =\sqrt{(4-2)^2+(7-5)^2+(8-7)^2} \\ & =\sqrt{(2)^2+(2)^2+(1)^2}=\sqrt{4+4+1} \end{aligned} $

$ A C=3 \Rightarrow A B: A C=6: 3=2: 1 $

Now, let say the internal bisector of the angle $A$ meets the side $B C$ that point

$ D\left(x_D, y_D, z_D\right) $

$\therefore x$-coordinate of point $D$,

$ x_D=\frac{m x_C+n x_B}{m+n}=\frac{2 \times 2+1 \times 2}{2+1} $

$ \Rightarrow x_D=2 $

$y$-coordinate of point $D$,

$ y_D=\frac{m y_C+n y_B}{m+n}=\frac{2 \times 5+1 \times 3}{2+1}=\frac{13}{3} $

$z$-coordinate of point $D$,

$ \begin{aligned} & z_D=\frac{m z_C+n z_B}{m+n}=\frac{2 \times 7+1 \times 4}{2+1}=\frac{18}{3}=6 \\ & \because \text { Point } D\left(2, \frac{13}{3}, 6\right) \end{aligned} $

So, length

$ \begin{aligned} & =\sqrt{(4-2)^2+\left(7-\frac{13}{3}\right)^2+(8-6)^2} \\ & =\sqrt{(2)^2+\left(\frac{8}{3}\right)^2+(2)^2} \\ & =\sqrt{4+\frac{64}{9}+4}=\sqrt{\frac{136}{9}}=\frac{2}{3} \sqrt{34} \end{aligned} $

Given, vector equation of plane

$ \begin{align*} & \mathbf{r}=(2+3 \lambda-\mu) \hat{\mathbf{i}}+(1-2 \lambda+3 \mu) \\ & \hat{\mathbf{j}}+(-2+2 \lambda+\mu) k \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{align*} $

Let $\mathbf{r}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

∴ Compare Eqs. (i) and (ii),

$ \begin{gather*} 2+3 \lambda-\mu=x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ 1-2 \lambda+3 \mu=y \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ -2+2 \lambda+\mu=z \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v) \end{gather*} $

Now, adding Eqs. (iii) and (iv),

$ \begin{align*} & 5 \lambda=x+z \\ & \Rightarrow \lambda=\frac{1}{5}(n+z) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi) \end{align*} $

On adding Eqs. (iv) and (vi)

$ \begin{align*} & & -1+4 \mu & =y+z \\ \Rightarrow & & \mu & =\frac{1}{4}(y+z+1) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vii) \end{align*} $

$ \begin{aligned} &\text { Putting value of } \lambda \text { and } \mu \text { in Eq. (iii), }\\ &\begin{aligned} & & 2+\frac{3}{5}(x+z)-\frac{1}{4}(y+z+1) & =x \\ & \Rightarrow & 40+12(x+z)-5(y+z+1) & =20 x \\ & \Rightarrow & 40+12 x+12 z-5 y-5 z-5 & =20 x \\ & \Rightarrow & -8 x-5 y+7 z+35 & =0 \\ & \Rightarrow & 8 x+5 y-7 z-35 & =0 \end{aligned} \end{aligned} $

Given projections of $A$ and $B$ are $P$ and $Q$ respectively

$ A(1,2,0) \text { and } B(2,1,1) $

Plane $P: x+y+z=3$

Vector perpendicular to plane $P$ : is

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

∴ vector equation of line $A P$ and $B Q$ are

$ \begin{gathered} A P: \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\lambda(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}), B Q: 2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\mathbf{k} \\ +\mu(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \end{gathered} $

∴ Any point on line $A P$ and $B Q$ is

$ P(1+\lambda, 2+\lambda, \lambda), Q(2+\mu, 1+\mu, 1+\mu) $

Point $P$ and $Q$ lies on the plane $x+y+z=3$

$ \therefore \quad 1+\lambda+2+\lambda+\lambda=3 \Rightarrow \lambda=0 $

and $2+\mu+1+\mu+1+\mu=3$

$ \Rightarrow \quad \mu=\frac{-1}{3} $

Point $P(1,2,0)$,

$ \begin{aligned} & Q\left(2-1-\left(\frac{-1}{3}\right), 1-\frac{1}{3}, 1-\frac{1}{3}\right) \\ & Q\left(\frac{5}{3}, \frac{2}{3}, \frac{2}{3}\right) \end{aligned} $

$ \begin{aligned} &\text { ∴ Distance between }\\ &\begin{aligned} P Q & =\sqrt{\left(\frac{5}{3}-1\right)^2+\left(\frac{2}{3}-2\right)^2+\left(\frac{2}{3}-0\right)^2} \\ P Q & =\sqrt{\frac{4}{9}+\frac{16}{9}+\frac{4}{9}} \Rightarrow P Q=\sqrt{\frac{24}{9}} \\ \Rightarrow P Q & =\frac{2 \sqrt{2}}{\sqrt{3}} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given points are }\\ &\begin{aligned} & A(4,3,2), B(5,4,6), C(-1,-1,5) \\ & \begin{aligned} A B & =\sqrt{(5-4)^2+(4-3)^2+(6-2)^2} \\ & =\sqrt{(1)^2+(1)^2+(4)^2}=\sqrt{18} \\ A B & =3 \sqrt{2} \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} A C & =\sqrt{(-1-4)^2+(-1-3)^2+(5-2)^2} \\ & =\sqrt{(-5)^2+(-4)^2+(3)^2} \\ & =\sqrt{25+16+9}=\sqrt{50} \\ A C & =5 \sqrt{2} \\ A B & : A C=3 \sqrt{2}: 5 \sqrt{2}=3: 5 \end{aligned} $

Let say the bisector of angle $A$ meet the side $B C$ that point $D\left(x_D, y_D, z_D\right)$ and $D$ divides $B C$ side in ratio 3: 5 internally

$ \begin{aligned} & x_D=\frac{m x_C+n x_B}{m+n}=\frac{3 \times(-1)+5 \times 5}{3+5}=\frac{22}{8} \\ & y_D=\frac{m y_C+n y_B}{m+n}=\frac{3 \times(-1)+5 \times(4)}{3+5}=\frac{17}{8} \\ & z_D=\frac{m z_C+n z_B}{m+n}=\frac{3 \times(5)+5 \times(6)}{3+5}=\frac{45}{8} \end{aligned} $

Therefore, coordinates of point $D$ is $\left(\frac{22}{8}, \frac{17}{8}, \frac{45}{8}\right)$.

Let the direction ratios of a line $L_1$ be $a_1, b_1, c_1$ and those of another line $L_2$ are $a_2, b_2, c_2$ and the angle $\theta$ between them is given by

$ \begin{equation*} \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}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

If both lines are parallel then, their direction ratios must be in equal proportion.

$i.e., \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} $

$ \begin{aligned} &\text { Now, from Eq. (i) }\\ &\begin{aligned} & \cos \theta=\frac{(2)\left(\frac{4}{\sqrt{19}}\right)+(5)\left(\frac{10}{\sqrt{19}}\right)+(7)\left(\frac{14}{\sqrt{19}}\right)}{\sqrt{2^2+5^2+7^2}} \\ & \sqrt{\begin{array}{c} \left(\frac{4}{\sqrt{19}}\right)^2+\left(\frac{10}{\sqrt{19}}\right)^2 \\ +\left(\frac{14}{\sqrt{19}}\right)^2 \end{array}} \\ & =\frac{\frac{8}{\sqrt{19}}+\frac{50}{\sqrt{19}}+\frac{98}{\sqrt{19}}}{\sqrt{78} \sqrt{\frac{312}{19}}}=\frac{\frac{106}{\sqrt{19}}}{\frac{78 \times 2}{\sqrt{19}}}=1 \\ & \Rightarrow \theta=0 \\ & \text { or } \frac{a_1}{a_2}=\frac{2}{\frac{4}{\sqrt{19}}}=\frac{\sqrt{19}}{2} \end{aligned} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \frac{b_1}{b_2}=\frac{5}{\frac{10}{\sqrt{19}}}=\frac{\sqrt{19}}{2} \\ & \frac{c_1}{c_2}=\frac{7}{\frac{14}{\sqrt{19}}}=\frac{\sqrt{19}}{2} \\ \because & \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \end{array} $

Hence, both lines are parallel to each other.

Given that, line

$ \frac{x-4}{1}=\frac{y-2}{1}=\frac{z-7}{2} $

Line is passing through point $P(4,+2,7)$ and direction ratios ( $1,1,2$ ) and lies on the given plane $a x+b y+z=7$

So, point will satisfy the equation of plane $a(4)+b(+2)+(7)=7$

$ \begin{equation*} 4 a+2 b=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Now direction ratios of the line and direction ratios of the plane will be perpendicular so their product must be zero

$ \begin{aligned} a_1 a_2+b_1 b_2 & +c_1 c_2 \\ & =(1)(a)+(1)(b)+(2)(1) \\ 0 & =a+b+2 \end{aligned} $

or    $ a+b=-2 $