The coordinates of the point,where the line through $A(3, 4, 1)$ and $B(5, 1, 6)$ crosses the $XZ$-plane,are

Let $P_1$ and $P_2$ be two planes given by $P_1: 10x + 15y + 12z - 60 = 0$ and $P_2: -2x + 5y + 4z - 20 = 0$. Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P_1$ and $P_2$?
$(A) \frac{x-1}{0} = \frac{y-1}{0} = \frac{z-1}{5}$
$(B) \frac{x-6}{-5} = \frac{y}{2} = \frac{z}{3}$
$(C) \frac{x}{-2} = \frac{y-4}{5} = \frac{z}{4}$
$(D) \frac{x}{1} = \frac{y-4}{-2} = \frac{z}{3}$

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

The equation of the plane containing the line of intersection of the planes $2x - y = 0$ and $y - 3z = 0$ and perpendicular to the plane $4x + 5y - 3z - 8 = 0$ is

If the angle between the line $2(x + 1) = y = z + 4$ and the plane $2x - \sqrt{\lambda} z + 4 = 0$ is $\frac{\pi}{6}$,then the value of $\lambda$ is

The angle $\theta$ between the line $x - 1 = \frac{y - 2}{-1} = \frac{z - 3}{2}$ and the plane $\vec{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 10$ is