⇒Line and Plane

Correct answer option is (A)

Given lines can be written as

$\frac{x-1}{-3}=\frac{y-2}{\frac{2p}{7}}=\frac{z-3}{2}\text{ and }\frac{x-1}{-\frac{3p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}$

As these lines are at right angles, we get

$(-3)(-\frac{3p}{7})+\left(\frac{2p}{7}\right)(1)+(2)(-5)=0$

$\begin{array}{rl}& \therefore \frac{9p}{7}+\frac{2p}{7}-10=0\\ & \therefore \frac{11p}{7}=10\Rightarrow p=\frac{70}{11}\end{array}$

Correct answer option is (B)

We have lines

$\frac{x-1}{-3}=\frac{7(y-2)}{2\lambda }=\frac{z-3}{2}\text{ and }\frac{7(1-x)}{3\lambda }=\frac{y-5}{1}=\frac{z-6}{-5}$

i.e. $\frac{x-1}{-3}=\frac{y-2}{\left(\frac{2\lambda }{7}\right)}=\frac{z-3}{2}$ and $\frac{x-1}{\left(\frac{-3\lambda }{7}\right)}=\frac{y-5}{1}=\frac{z-6}{-5}$

Since given lines are at right angles, we write

$\begin{array}{rl}& (-3)\left(\frac{-3\lambda }{7}\right)+\left(\frac{2\lambda }{7}\right)(1)+(2)(-5)=0\\ & \therefore \frac{9\lambda }{7}+\frac{2\lambda }{7}-10=0\Rightarrow 11\lambda =70\Rightarrow \lambda =\frac{70}{11}\end{array}$

Correct answer option is (C)

Direction ratios of a line joining $(3,1,4)$ and $(7,2,12)$ are $4,1,8$

Let $(a_1,b_1,c_1)=(2,2,1)$ and $(a_2,b_2,c_2)=(4,1,8)$.

Hence angle $\theta$ between the lines is given by

$\begin{array}{rl}\cos \theta & =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1}\sqrt{a_2^2+b_2^2+c_2^2}}\\ & =\frac{8+2+8}{\sqrt{4+4+1}\cdot \sqrt{16+1+64}}=\frac{18}{\sqrt{9}\cdot \sqrt{81}}=\frac{18}{(3)(9)}=\frac{2}{3}\\ \therefore \theta & ={\cos }^{-1}\left(\frac{2}{3}\right)\end{array}$