Let a line $L$ pass through the point $P(2, 3, 1)$ and be parallel to the line $x + 3y - 2z - 2 = 0 = x - y + 2z$. If the distance of $L$ from the point $(5, 3, 8)$ is $\alpha$,then $3\alpha^2$ is equal to $......$.

Let $L_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $L_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in R$,be two lines,which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A(1,1,-1)$ on $L_2$,then the value of $26 \alpha(PB)^2$ is . . . . . . .

If the lines $\frac{1-x}{3}=\frac{7y-14}{2\lambda}=\frac{z-3}{2}$ and $\frac{7-7x}{3\lambda}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles,then $\lambda=$

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

The distance of the point $(2, 3, 4)$ from the line $1 - x = \frac{y}{2} = \frac{1}{3}(1 + z)$ is

If the lines $\frac{1-x}{2}=\frac{7y+4}{2\lambda}=\frac{2z-5}{2}$ and $\frac{7-7x}{3\lambda}=\frac{y-1}{7}=\frac{6-z}{5}$ are at right angles,then the value of $\lambda$ is