The equation of the plane passing through the point $(2,-1,-3)$ and parallel to the lines $\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$ is

The coordinates of the point where the line joining $(1, 1, 1)$ and $(2, 2, 2)$ intersects the plane $x + y + z = 9$ are:

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.

The plane $3x + 4y + 6z + 7 = 0$ is rotated about the line $r = (\hat{i} + 2\hat{j} - 3\hat{k}) + t(2\hat{i} - 3\hat{j} + \hat{k})$ until the plane passes through the origin. The equation of the plane in the new position is

Find the ratio in which the plane $x - 2y + 3z = 17$ divides the line segment joining the points $(-2, 4, 7)$ and $(3, -5, 8)$.

If $\hat{i}+\hat{j}, \hat{j}+\hat{k}, \hat{k}+\hat{i}, \hat{i}-\hat{j}, \hat{j}-\hat{k}$ are the position vectors of the points $A, B, C, D, E$ respectively, then the point of intersection of the line $AB$ and the plane passing through $C, D, E$ is