The vertices $B$ and $C$ of a $\Delta ABC$ lie on the line $\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}$ such that $BC = 5 \text{ units}$. If the vertex $A$ is $(1, -1, 2)$,then the area of this triangle (in $\text{sq. units}$) is:

If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$,then the angle between the lines is

Let the area of the triangle formed by the lines $x+2=y-1=z$,$\frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}$ and $\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}$ be $A$. Then $A^2$ is equal to . . . . . . .

Let $P$ be the foot of the perpendicular from the point $A(1, 2, 2)$ on the line $L: \frac{x-1}{1} = \frac{y+1}{-1} = \frac{z-2}{2}$. Let the line $\overrightarrow{r} = (-\hat{i} + \hat{j} - 2\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$,$\lambda \in R$,intersect the line $L$ at $Q$. Then $2(PQ)^2$ is equal to:

Find the vector equation for the line passing through the points $(-1, 0, 2)$ and $(3, 4, 6).$

The coordinates of the foot of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{6 - x}{-3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ are: