The distance of the point $(-2, 4, -5)$ from the line $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$ is

Let a straight line $L$ pass through the point $P(2, -1, 3)$ and be perpendicular to the lines $\frac{x-1}{2} = \frac{y+1}{1} = \frac{z-3}{-2}$ and $\frac{x-3}{1} = \frac{y-2}{3} = \frac{z+2}{4}$. If the line $L$ intersects the $yz$-plane at the point $Q$,then the distance between the points $P$ and $Q$ is:

Let the point $A$ be the foot of the perpendicular drawn from the point $P(a, b, 0)$ to the line $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3}$. If the midpoint of the line segment $PA$ is $(0, \frac{3}{4}, -\frac{1}{4})$,then the value of $a^2+b^2+\alpha^2$ is equal to:

Find the coordinates of the point where the line passing through $(5, 1, 6)$ and $(3, 4, 1)$ crosses the $YZ$-plane.

Let $P$ be the point of intersection of the lines $\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$ and $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$. Then,the shortest distance of $P$ from the line $4x=2y=z$ is

One vertex of a rectangular parallelepiped is at the origin $O$ and the lengths of its edges along $x, y$ and $z$ axes are $3, 4$ and $5$ units respectively. Let $P$ be the vertex $(3, 4, 5)$. Then the shortest distance between the diagonal $OP$ and an edge parallel to the $z$-axis,not passing through $O$ or $P$,is: