If the distance between the plane $ax - 2y + z = k$ and the plane containing the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$ is $\sqrt{6}$, then $|k|$ is

The image of the point $(5, 2, 6)$ with respect to the plane $x + y + z = 9$ is

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

Let a plane $P$ contain two lines $\overrightarrow{r} = \hat{i} + \lambda(\hat{i} + \hat{j}), \lambda \in R$ and $\overrightarrow{r} = -\hat{j} + \mu(\hat{j} - \hat{k}), \mu \in R$. If $Q(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from the point $M(1, 0, 1)$ to $P$,then $3(\alpha + \beta + \gamma)$ equals

If the line $\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}$ is parallel to the plane $ax + by + cz + d = 0$,then:

The shortest distance between the $z$-axis and the line $x + y + 2z - 3 = 0 = 2x + 3y + 4z - 4$ is