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

Find the shortest distance between the lines whose vector equations are $\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}$ and $\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}$.

If the lines $\frac{x+1}{-10}=\frac{y+k}{-1}=\frac{z-4}{1}$ and $\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}$ intersect each other,then the value of $k$ is

The equation of a line passing through the point $(-3, 2, -4)$ and equally inclined to the axes is:

The Cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} .$ Write its vector form.

Consider the lines $L_1: x-1=y-2=z$ and $L_2: x-2=y=z-1$. Let the feet of the perpendiculars from the point $P(5,1,-3)$ on the lines $L_1$ and $L_2$ be $Q$ and $R$ respectively. If the area of the triangle $PQR$ is $A$,then $4A^2$ is equal to: