The position vector of the point where the line $r = i - j + k + t(i + j - k)$ meets the plane $r \cdot (i + j + k) = 5$ is

$A$ plane is making intercepts $2, 3, 4$ on $X, Y$ and $Z$-axes respectively. Another plane is passing through the point $(-1, 6, 2)$ and is perpendicular to the line joining the points $(1, 2, 3)$ and $(-2, 3, 4)$. Then the angle between the two planes is

If the lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar,then $k$ can have:

The equation of the plane containing the line $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-4}{-2}$ and the point $(0,5,0)$ is

If the line $\frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z + 4}{3}$ lies in the plane $lx + my - z = 9$,then $l^2 + m^2 = \dots$

The image of the point with position vector $\vec{i} + 3\vec{k}$ in the plane $\vec{r} \cdot (\vec{i} + \vec{j} + \vec{k}) = 1$ is