Line $L$ passes through two points $(2, -3, 1)$ and $(3, -4, -5)$. If point $(0, a, b)$ lies on the line $L$,then $a+b =$ . . . . . . .

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

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

The length of the perpendicular drawn from the origin to the line $\bar{r} = (4\hat{i} + 2\hat{j} + 4\hat{k}) + \lambda(3\hat{i} + 4\hat{j} - 5\hat{k})$ is .......

The angle between the straight lines $\frac{x - 2}{2} = \frac{y - 1}{5} = \frac{z + 3}{-3}$ and $\frac{x + 1}{-1} = \frac{y - 4}{8} = \frac{z - 5}{4}$ is