What is the point of intersection of the line joining the points $(3, 4, 1)$ and $(5, 1, 6)$ with the $xy$-plane?

The equation of a line passing through the point $(2, 4, 6)$ and parallel to the line $3x + 4 = 4y - 1 = 1 - 4z$ is

Let $L$ be a line passing through a point $A$ and parallel to the vector $2 \hat{i}+\hat{j}-2 \hat{k}$. Let $-7 \hat{i}-5 \hat{j}+11 \hat{k}$ be the position vector of a point $P$ on $L$ such that $|\overline{AP}|=12$. Then the position vector of $A$ can be

Lines $\frac{1-x}{3}=\frac{y-2}{1}=\frac{z-1}{2}$ and $\frac{x-2}{p}=\frac{y-1}{2}=\frac{z-2}{1}$ are mutually perpendicular to each other,then $p=$ . . . . . . .

Find the equation of the line passing through the point $P(2, -3, 1)$ and perpendicular to the line $L: \frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{-1}$.

The vertices $B$ and $C$ of a triangle $ABC$ lie on the line $\frac{x}{1}=\frac{1-y}{-2}=\frac{z-2}{3}$. The coordinates of $A$ and $B$ are $(1, 6, 3)$ and $(4, 9, \alpha)$ respectively, and $C$ is at a distance of $10$ units from $B$. The area (in sq. units) of $\Delta ABC$ is: