The three different face diagonals of a cuboid (rectangular parallelepiped) have lengths $39, 40, 41$. The length of the main diagonal of the cuboid which joins a pair of opposite corners is

The edge of a cube is of length $a$. Then the shortest distance between the diagonal of a cube and an edge skew to it is:

The angle between two diagonals of a cube is

$\Pi_1, \Pi_2, \Pi_3$ are three planes which are respectively parallel to the $YZ, ZX$ and $XY$ planes at distances $a, b$ and $c$ forming a rectangular parallelopiped. $d_1$ is a diagonal of the face of $XY$-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

Find the distance of the point $P(3, 8, 2)$ from the line $\frac{x-1}{2} = \frac{y-3}{4} = \frac{z-2}{3}$ measured parallel to the plane $3x + 2y - 2z + 15 = 0$.

Find the image of the point $(1, 2, 3)$ in the line $\vec{r} = (6\hat{i} + 7\hat{j} + 7\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 2\hat{k})$.