If a line $L$ is common to the planes $x-y+z+2=0$ and $2 x+y-2 z+5=0$ then the direction cosines of the line $L$ are

Let the foot of the perpendicular drawn from the point $(1,2,3)$ to a plane be $(-1,3,-2)$. Then, the perpendicular distance from the origin to the plane is

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

$\Pi_1, \Pi_2, \Pi_3$ are three planes which are respectively parallel to the $Y Z, Z X$ and $X Y$ planes at distances $a, b$ and $c$ forming a rectangular parallelopiped. $d_1$ is a diagonal of the face of $X Y$-plane not passing through the origin and $d_2$ is a diagonal of the plane $\Pi_2$ coterminous with $d_1$. If none of the coordinates of the vertices of the parallelopiped are negative, then the angle between $d_1$ and $d_2$ is