Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $5y + 8 = 0$.

If a plane $P$ passes through the points $(1,0,0)$ and $(0,1,0)$ and makes an angle $\frac{\pi}{4}$ with the plane $x+y=3$, then the direction ratios of a normal to that plane $P$ are

The equation of the plane passing through the intersection of the planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to the $x$-axis is:

Let the coordinates of one vertex of $\triangle ABC$ be $A(0, 2, \alpha)$ and the other two vertices lie on the line $\frac{x+\alpha}{5} = \frac{y-1}{2} = \frac{z+4}{3}$. For $\alpha \in \mathbb{Z}$,if the area of $\triangle ABC$ is $21$ sq. units and the line segment $BC$ has length $2\sqrt{21}$ units,then $\alpha^2$ is equal to $...........$.

The image of the point $(-1, 3, 4)$ in the plane $x - 2y = 0$ is

The image of the line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ in the plane $2x-y+z+3=0$ is the line