The shortest distance between the line $r = 2\hat{i} - 2\hat{j} + 3\hat{k} + \lambda(\hat{i} - \hat{j} + 4\hat{k})$ and the plane $r \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$ is

The line passing through $(1, 1, -1)$ and parallel to the vector $\hat{i} + 2 \hat{j} - \hat{k}$ meets the line $\frac{x - 3}{-1} = \frac{y + 2}{5} = \frac{z - 2}{-4}$ at $A$ and the plane $2 x - y + 2 z + 7 = 0$ at $B$. Then $AB = $

If the three planes $x = 5$,$2x - 5ay + 3z - 2 = 0$,and $3bx + y - 3z = 0$ pass through a common line,then the value of $(a, b)$ is:

The point of intersection of the line $\frac{x - 1}{3} = \frac{y + 2}{4} = \frac{z - 3}{-2}$ and the plane $2x - y + 3z - 1 = 0$ is:

$A$ plane which passes through the point $(3, 2, 0)$ and the line $\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}$ is

$A$ line $l$ passing through the origin is perpendicular to the lines $l_{1}: \overrightarrow{r}=(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k}$ and $l_{2}: \overrightarrow{r}=(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k}$. If the coordinates of the point in the first octant on $l_{2}$ at a distance of $\sqrt{17}$ from the point of intersection of $l$ and $l_{1}$ are $(a, b, c)$,then $18(a+b+c)$ is equal to ........ .