The coordinates of the point where the line $\frac{x - 6}{-1} = \frac{y + 1}{0} = \frac{z + 3}{4}$ meets the plane $x + y - z = 3$ are

Find the equation of the plane which is perpendicular to the plane $5x + 3y + 6z + 8 = 0$ and which contains the line of intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$.

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

The angle made by the vector $2\hat{i}-\hat{j}+\hat{k}$ with the plane represented by $\vec{r} \cdot(\hat{i}+\hat{j}+2\hat{k})=7$ is (in $^{\circ}$)

$A$ line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. The length of the line segment $PQ$ is equal to:

The plane $4x + 7y + 4z + 81 = 0$ is rotated through a right angle about its line of intersection with the plane $5x + 3y + 10z = 25$. The equation of the plane in its new position is $x - 4y + 6z = k$,where $k$ is: