The acute angle $\theta$ between the lines $2x = 3y = -z$ and $6x = -y = -4z$ is:

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:

The straight line $\frac{x - 3}{3} = \frac{y - 2}{1} = \frac{z - 1}{0}$ is

The vector equation of the line whose Cartesian equations are $y=2$ and $4x-3z+5=0$ is

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 parametric equations of a line passing through the points $A(3, 4, -7)$ and $B(1, -1, 6)$ are