$A$ line passing through the points $P(1, -1, 2)$ and $Q(2, 0, 1)$ meets the $XY$-plane and the $YZ$-plane at points $A$ and $B$ respectively. The distance $AB$ is equal to...

Show that the line joining the origin to the point $(2,1,1)$ is perpendicular to the line determined by the points $(3,5,-1)$ and $(4,3,-1).$

The shortest distance between the lines $r = (3t - 4)\hat{i} - 2\hat{j} - (1 + 2t)\hat{k}$ and $r = (6 + s)\hat{i} + (2 - 2s)\hat{j} + 2(1 + s)\hat{k}$ is

If a point $R(4, y, z)$ lies on the line segment joining the points $P(2, -3, 4)$ and $Q(8, 0, 10)$,then the distance of $R$ from the origin is

Let $(\alpha, \beta, \gamma)$ be the coordinates of the foot of the perpendicular drawn from the point $(5, 4, 2)$ on the line $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$. Then the length of the projection of the vector $\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$ on the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$ is:

If $A(1,2,3), B(3,7,-2), C(6,7,7)$ and $D(-1,0,-1)$ are points in a plane, then the vector equation of the line passing through the centroids of $\triangle ABD$ and $\triangle ACD$ is