The lines $\frac{6x-6}{18} = \frac{y+1}{3} = \frac{z-1}{5}$ and $\frac{3x+6}{12} = \frac{y-1}{3} = \frac{z+1}{2}$ are $\dots$

Find the angle between the lines $\frac{x-2}{3} = \frac{y+1}{-2}, z=2$ and $\frac{x-1}{1} = \frac{2y+3}{3} = \frac{z+5}{2}$.

Let $A$ be the point of intersection of the lines $L_1: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}$ and $L_2: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$. Let $B$ and $C$ be points on the lines $L_1$ and $L_2$ respectively such that $AB = AC = \sqrt{15}$. Then the square of the area of the triangle $ABC$ is:

Find the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$.

If the shortest distance between the lines $\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$,then the sum of all possible values of $\lambda$ is :

The line $L_1$ is parallel to the vector $\vec{a} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and passes through the point $(7, 6, 2)$,and the line $L_2$ is parallel to the vector $\vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and passes through the point $(5, 3, 4)$. The shortest distance between the lines $L_1$ and $L_2$ is: