The distance between the point with position vector $-\hat{i}-5 \hat{j}-10 \hat{k}$ and the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ with the plane $x-y+z=5$, is $\_\_\_\_$ (units)

The equation of the straight line through the origin and parallel to the line

$$ \begin{aligned} & (b+c) x+(c+a) y+(a+b) z=k= \\ & (b-c) x+(c-a) y+(a-b) z \text { is } \end{aligned} $$

If the distance between the planes $8 x+12 y-14 z=2$ and $4 x+6 y-7 z=2$ can be expressed as $\frac{1}{\sqrt{N}}$, then the value of $\frac{N(N+1)}{2}$ is

The value of ' $a$ ' for which the lines $\frac{x-2}{1}=\frac{y-9}{2}=\frac{z-13}{3}$ and $\frac{x-a}{-1}=\frac{y-7}{2}$ $=\frac{z+2}{-3}$ intersect, is