If the points $A(-1,3,2)$, $B(-4,2,-2)$ and $C(5,5,\lambda)$ are collinear, then value of $\lambda$ is :

Equation of a line passing through (1, 2, -3) which is parallel to the line $\frac{x+2}{3} = \frac{y-3}{-4} = \frac{z+5}{5}$ is

Distance between two planes $x + 2y - z = 5$ and $2x + 4y - 2z + 2 = 0$ is

The shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{4} = \frac{y-4}{6} = \frac{z-5}{8}$ is equal to

The correct order of steps from A to E, for finding value of p, so that the lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are at right angle is:

A. $p = \frac{70}{11}$

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

C. $\frac{x-1}{-3} = \frac{y-2}{2\frac{p}{7}} = \frac{z-3}{2}$, $\frac{x-1}{-\frac{3p}{7}} = \frac{y-5}{1} = \frac{z-6}{-5}$

D. $\frac{9p}{7} + \frac{2p}{7} - 10 = 0$

E. $\frac{11p}{7} = 10$

Choose the correct answer from the options given below:

If the lines $\frac{1-x}{3} = \frac{3y-6}{k} = \frac{3-z}{-2}$ and $\frac{1-x}{2k} = \frac{y-5}{3} = \frac{6-z}{5}$ are perpendicular to each other, then $k$ is equal to

The angle between the lines $\vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})$ and $\vec{r} = 5\hat{j} - 2\hat{k} + \mu(3\hat{i} + 2\hat{j} + 6\hat{k})$ is :