The square of the distance of the point $P(5,6,7)$ from the line $\frac{x-2}{2} = \frac{y-5}{3} = \frac{z-2}{4}$ is equal to:

If the distance of point $B(2, 1, -3)$ from the line passing through the point $A(4, -2, 2)$ and parallel to the vector $\vec{c} = -4\hat{i} - 6\hat{j} - 2\hat{k}$ is $x$, then $x^4 + x^2 + 541 =$

The length of the perpendicular drawn from the point $(3, -1, 11)$ to the line $\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ is

The shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$ is (in $\sqrt{3}$)

The point $P$ lies on the line $AB$,where $A \equiv (2, 4, 5)$ and $B \equiv (1, 2, 3)$. If the $z$-coordinate of point $P$ is $3$,then its $y$-coordinate is:

If the line passing through the points $(-5, 1, 3)$ and $(1, 2, 0)$ is perpendicular to the line passing through the points $(x, 2, 1)$ and $(0, -4, 6)$,then $x = \dots$