The line joining points $(3, 5, -7)$ and $(-2, 1, 8)$ meets the $yz$-plane at which point?

The distance from the point $-i + 2j + 6k$ to the straight line passing through the point $(2, 3, -4)$ and parallel to the vector $6i + 3j - 4k$ is

If the shortest distance between the lines $\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$ and $\frac{x-\lambda}{3}=\frac{y-2\sqrt{6}}{4}=\frac{z+2\sqrt{6}}{5}$ is $6$,then the square of the sum of all possible values of $\lambda$ is

Show that the line through the points $(1, -1, 2)$ and $(3, 4, -2)$ is perpendicular to the line through the points $(0, 3, 2)$ and $(3, 5, 6)$.

Let the area of the triangle formed by the lines $x+2=y-1=z$,$\frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}$ and $\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}$ be $A$. Then $A^2$ is equal to . . . . . . .

The line $L$ given by $\frac{x - 2}{2} = \frac{y - 1}{b} = \frac{z + 1}{c}$ passes through the point $(1, 2, 3)$. Another line $K$ is parallel to line $L$ and has the equation $\frac{x + 2}{a} = \frac{y - 3}{2} = \frac{z + 4}{d}$. Then the distance between line $L$ and $K$ is