Find the point of intersection of the line $\frac{x}{1} = \frac{y}{2} = \frac{z}{2}$ and the plane $2x + y + z = 6$.

If the foot of the perpendicular from the point $A(-1, 4, 3)$ on the plane $P: 2x + my + nz = 4$ is $B\left(-2, \frac{7}{2}, \frac{3}{2}\right)$,then the distance of the point $A$ from the plane $P$,measured parallel to a line with direction ratios $3, -1, -4$,is equal to.

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:

Find the equation of the line perpendicular to the plane $2x + 4y - 5z = 10$ passing through the origin.

The line of intersection of the planes $\vec{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1$ and $\vec{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2$ is parallel to which vector?

Find the coordinates of the point of intersection of the line $\frac{x - 6}{-1} = \frac{y + 1}{0} = \frac{z + 3}{4}$ and the plane $x + y - z = 3$.