Find the distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$.

If the line $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$ lies on the plane $2x - 4y + z = 7$,then $k = . . . . $

For a positive real number $p$,if the perpendicular distance from a point $-\hat{i} + p\hat{j} - 3\hat{k}$ to the plane $\vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = 7$ is $6$ units,then $p=$

If $L$ is a line common to the planes $3x + 4y + 7z = 1$ and $x - y + z = 5$, then the direction ratios of the line $L$ are:

Find the equation of the plane containing the point $(0, 7, -7)$ and the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$.

If the plane $P$ passes through the intersection of two mutually perpendicular planes $2x + ky - 5z = 1$ and $3kx - ky + z = 5$,where $k < 3$,and intercepts a unit length on the positive $x$-axis,then the intercept made by the plane $P$ on the $y$-axis is