Let a line having direction ratios $1, -4, 2$ intersect the lines $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}$ and $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}$ at the points $A$ and $B$ respectively. Then $( AB )^{2}$ is equal to

Find the shortest distance between the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2\hat{i} - \hat{j} - \hat{k})$ and $\vec{r} = (\hat{i} - \hat{j} - \hat{k}) + \mu(\hat{i})$.

Find the Cartesian equation of the line passing through the point $\hat{i} + 2\hat{j} + 2\hat{k}$ and parallel to the line joining the points $2\hat{i} - \hat{j} + \hat{k}$ and $-\hat{i} + 4\hat{j} + \hat{k}$.

The point of intersection of the lines $l_1: r(t) = (i - 6j + 2k) + t(i + 2j + k)$ and $l_2: R(u) = (4j + k) + u(2i + j + 2k)$ is

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3}$ is

Let $P$ be the foot of the perpendicular from the point $A(1, 2, 2)$ on the line $L: \frac{x-1}{1} = \frac{y+1}{-1} = \frac{z-2}{2}$. Let the line $\overrightarrow{r} = (-\hat{i} + \hat{j} - 2\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$,$\lambda \in R$,intersect the line $L$ at $Q$. Then $2(PQ)^2$ is equal to: