If ${L_1}$ is the line of intersection of the planes $2x - 2y + 3z - 2 = 0$ and $x - y + z + 1 = 0$,and ${L_2}$ is the line of intersection of the planes $x + 2y - z - 3 = 0$ and $3x - y + 2z - 1 = 0$,then the distance of the origin from the plane containing the lines ${L_1}$ and ${L_2}$ is:

The reflection of the point $(-1, 3, 4)$ with respect to the plane $x - 2y = 0$ is .....

The perpendicular distance from the origin to the plane containing the two lines,$\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}$ and $\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}$,is

Let the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$ intersect the plane containing the lines $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$ and $4ax-y+5z-7a=0=2x-5y-z-3, a \in R$ at the point $P(\alpha, \beta, \gamma)$. Then the value of $\alpha+\beta+\gamma$ equals...

Find the length of the perpendicular from the point $(7, 14, 5)$ to the plane $2x + 4y - z = 2$ and the coordinates of the foot of the perpendicular.

$A$ 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 ........ .