The point of intersection of lines $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1}$ and $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ is

If the lines $2x = ky = -z$ and $6x = -y = -4z$ are perpendicular to each other, then the value of $k$ is:

Let a line passing through the point $(-1, 2, 3)$ intersect the lines $L_1: \frac{x-1}{3} = \frac{y-2}{2} = \frac{z+1}{-2}$ at $M(\alpha, \beta, \gamma)$ and $L_2: \frac{x+2}{-3} = \frac{y-2}{-2} = \frac{z-1}{4}$ at $N(a, b, c)$. Then the value of $\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}$ equals

Find the vector equation of the line which is parallel to the vector $3 \hat{i}-2 \hat{j}+6 \hat{k}$ and which passes through the point $(1, -2, 3)$.

The shortest distance between the skew lines $r = (-\hat{i} + 3\hat{k}) + t(2\hat{i} + 3\hat{j} + 6\hat{k})$ and $r = (3\hat{i} + \hat{j} - \hat{k}) + s(2\hat{i} - \hat{j} + 2\hat{k})$ is

The equation of the line passing through the point $(3,1,2)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$ is