The distance of the point $(1, -2, 4)$ from the plane passing through the point $(1, 2, 2)$ and perpendicular to the planes $x - y + 2z = 3$ and $2x - 2y + z + 12 = 0$ is:

Find the equation of the plane containing the lines $\vec{r} = (\hat{i} + \hat{j}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})$ and $\vec{r} = (\hat{i} + \hat{j}) + \mu(-\hat{i} + \hat{j} - 2\hat{k})$.

$A$ perpendicular is drawn from a point $P$ on the line $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{1}$ to the plane $x + y + z = 3$ such that the foot of the perpendicular $Q$ also lies on the plane $x - y + z = 3$. Then the coordinates of $Q$ are

If the line $\frac{2x - 8}{\sin \beta} = \frac{y - \sin \alpha}{1} = \frac{z - 1}{\cos \alpha}$,where $\beta \in R$ and $\sin \beta \neq 1$,lies in the plane $2x - (\sin \beta)y + (\cos \beta)z = k$ for all $\alpha \in R$,then:

The equation of the plane passing through the line $\frac{x - 1}{5} = \frac{y + 2}{6} = \frac{z - 3}{4}$ and the point $(4, 3, 7)$ is

What is the distance between the line $r = (2i - 2j + 3k) + \lambda (i - j + 4k)$ and the plane $r \cdot (i + 5j + k) = 5$?