Three Dimensional Geometry

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

$ \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 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 $P=(1,-2,-2)$

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 $

Given, lines

$ \begin{aligned} \mathbf{r} & =(-\hat{\mathbf{i}}+3 \hat{\mathbf{k}})+t(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \\ \text { and } \mathbf{r} & =(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+s(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{aligned} $

Shortest distance between lines is

$ \begin{aligned} & ((3+1) \hat{\mathbf{i}}+\hat{\mathbf{j}}-(\hat{\mathbf{k}}+3) \hat{\mathbf{k}}) \\ S D & =\frac{-[(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \times(2 \hat{\mathbf{i}}-\mathbf{j}+2 \mathbf{k})]}{|(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})| \times(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \mid} \\ S D & =\frac{(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-4 \hat{\mathbf{k}}) \cdot(12 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}-8 \hat{\mathbf{k}})}{|12 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}|} \\ S D & =\frac{48+8+32}{\sqrt{144+64+64}}=\frac{88}{4 \sqrt{17}}=\frac{22}{\sqrt{17}} \end{aligned} $

Direction ratio of diagonal $d_1$ i.e., direction of $A E$ is $a \hat{\mathbf{i}}-b \hat{\mathbf{j}}$

Direction ratio of diagonal $d_2$ i.e., direction of $A D a \hat{\mathbf{i}}-c \hat{\mathbf{k}}$

Angle between diagonal $d_1$ and diagonal $d_2$

$ \begin{aligned} \cos \theta & =\frac{(a \hat{\mathbf{i}}-b \hat{\mathbf{j}}) \cdot(a \hat{\mathbf{i}}-c \hat{\mathbf{k}})}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}} \\ & =\frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}} \\ \Rightarrow \quad \theta & =\cos ^{-1}\left(\frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}}\right) \end{aligned} $

$ \begin{aligned} & \text {We have, } a+b+c=0 \\ & \text { and } 2 a b+2 a c-b c=0 \\ & \quad a=-(b+c) \\ & \therefore \quad-2(b+c) b+-2(b+c) c-b c=0 \\ & \Rightarrow \quad-2 b^2-2 b c-2 b c-2 c^2-b c=0 \\ & \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad 2 b^2+5 b c+2 c^2=0 \\ & \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad 2 b^2+4 b c+b c+2 c^2=0 \\ & \Rightarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \quad(2 b+c)(b+2 c)=0 \\ & \Rightarrow \quad b=-2 c, b=-c / 2 \\ & \quad a=-(-2 c+c)=c \\ & \Rightarrow \quad a=-\left(\frac{-c}{2}+c\right)=\frac{-c}{2} \\ & \therefore a_1=1, b_1=-2, c_1=1, \\ & \quad a_2=\frac{-1}{2}, b_2=\frac{-1}{2}, c_2=1 \\ & \quad a_1=1, b_1=-2, c_1=1, \\ & a_2=1, b_2=1, c_2=-2 \\ & \cos \theta=\frac{(1)(1)+(-2)(1)+(1)(-2)}{\sqrt{1+4+1} \sqrt{1+1+4}} \\ & \cos \theta=\frac{-1}{2} \Rightarrow \theta=\frac{2 \pi}{3} \end{aligned} $

Let the equation of plane is, $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

where $a, b, c$ are $x, y$ and $z$ intercepts ∴ Centroid of $\triangle A B C$ is $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$.

Given, $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)=(2,3,5)$

$ \therefore \quad a=6, b=9, c=15 $

Equation of plane is, $\frac{x}{6}+\frac{y}{9}+\frac{z}{15}=1$

$ 15 x+10 y+6 z=90 $

Equation of plane ( $\pi$ ) containing the points

$ \begin{aligned} & (0,-5,-1),(1,-2,5),(-3,5,0) \\ & \text { is }\left|\begin{array}{ccc} x-0 & y+5 & z+1 \\ 1-0 & -2+5 & 5+1 \\ -3-0 & 5+5 & 0+1 \end{array}\right|=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc} x & y+5 & z+1 \\ 1 & 3 & 6 \\ -3 & 10 & 1 \end{array}\right|=0 \\ & \Rightarrow \quad 3 x+y-z+4=0 \\ & \Rightarrow \quad \mathbf{r}(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+4=0 \\ & \Rightarrow \quad \mathbf{r}(-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=4 \\ & \Rightarrow \mathbf{r}\left(\frac{-3}{\sqrt{11}}-\frac{\hat{\mathbf{j}}}{\sqrt{11}}+\frac{\hat{\mathbf{k}}}{\sqrt{11}}\right)=\frac{4}{\sqrt{11}} \end{aligned} $

∴ Unit normal vector to plane $\pi$ is

$ \hat{\mathbf{n}}=\frac{-3}{\sqrt{11}} \hat{\mathbf{i}}-\frac{\hat{\mathbf{j}}}{\sqrt{11}}+\frac{\hat{\mathbf{k}}}{\sqrt{11}} $

∴ length of the projection of the unit normal vector $\hat{\mathbf{n}}$ to plane $\pi$ on the line $L$ is

$ \left|\frac{\left(\frac{-3}{\sqrt{11}} \hat{\mathbf{i}} \frac{-\hat{\mathbf{j}}}{\sqrt{11}} \frac{+\hat{\mathbf{k}}}{\sqrt{11}}\right) \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}-\hat{\mathbf{k}})}{\sqrt{1+25+36}}\right| $

$ =\left|\frac{\frac{-3}{\sqrt{11}} \frac{-5}{\sqrt{11}} \frac{-6}{\sqrt{11}}}{\sqrt{62}}\right|=\left|\frac{-14}{\sqrt{11} \sqrt{62}}\right|=\frac{14}{\sqrt{682}} $

Equation of line passing through the points $(a, 2,-4)$ and $(5,3, b)$ is given by

$ \begin{aligned} \frac{x-a}{5-a} & =\frac{y-2}{3-2}=\frac{z+4}{b+4} \\ \Rightarrow \quad \frac{x-a}{5-a} & =\frac{y-2}{1}=\frac{z+4}{b+4} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

The equation of $Z X$-plane, $y=0$ Since, line (i) meets the $Z X$-plane.

$ \begin{array}{ll} \Rightarrow & \frac{x-a}{5-a}=\frac{0-2}{1}=\frac{z+4}{b+4} \\ \Rightarrow & \frac{x-a}{5-a}=-2 \text { and } \frac{z+4}{b+4}=-2 \\ \Rightarrow & x-a=-2(5-a) \\ \text { and } & z+4=-2(b+4) \end{array} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad x=a-10+2 a \\ & \text { and } z+4=-2 b-8 \\ & \Rightarrow \quad x=3 a-10 \text { and } z=-2 b-12 \\ & \text { and } y=0 \\ & \therefore \quad 3 a-10=-a+2 b \text { and } \\ & -2 b-12=a+b \\ & \Rightarrow \quad 4 a-2 b=10 \text { and } a+3 b=-12 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \Rightarrow \quad 2 a-b=5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned}\\ &\text { On solving Eqs (i) and (ii), we get }\\ &\begin{aligned} a=\frac{3}{7} & \text { and } b=\frac{-29}{7} \\ \therefore \quad 14 a+7 b & =14\left(\frac{3}{7}\right)+7\left(\frac{-29}{7}\right) \\ & =6-29=-23 \end{aligned} \end{aligned} $