The straight lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x - 1}{2} = \frac{y - 2}{2} = \frac{z - 3}{-2}$ are

If the shortest distance between the lines $r=(3 \hat{i}+4 \hat{j}-2 \hat{k})+t(-\hat{i}+2 \hat{j}+\hat{k})$ and $r=(\hat{i}-7 \hat{j}-2 \hat{k})+s(\hat{i}+3 \hat{j}+2 \hat{k})$ is equivalent to the projection of $P=-2 \hat{i}+11 \hat{j}$ on $Q$, then a possible vector $Q$ is

The equation of a line passing through the point $(2, -1, 1)$ and parallel to the line joining the points $\hat{i} + 2\hat{j} + 2\hat{k}$ and $-\hat{i} + 4\hat{j} + \hat{k}$ is

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is

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

Let $P$ and $Q$ be the points on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ which are at a distance of $6$ units from the point $R(1,2,3)$. If the centroid of the triangle $PQR$ is $(\alpha, \beta, \gamma)$,then $\alpha^2+\beta^2+\gamma^2$ is: