If the lines $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}$ and $\frac{x-3}{1} = \frac{y-k}{2} = \frac{z}{1}$ intersect,then find the value of $k$.

If $A(-2, 4, a)$,$B(1, b, 3)$,$C(c, 0, 4)$,and $D(-5, 6, 1)$ are collinear points,then $a+b+c=$

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:

$\triangle ABC$ is formed by $A(1, 8, 4)$,$B(0, -11, 4)$ and $C(2, -3, 1)$. If $D$ is the foot of the perpendicular from $A$ to $BC$,then the coordinates of $D$ are

The angle between the two lines $\frac{x-3}{1}=\frac{y-2}{2}=\frac{z+4}{2}$ and $\frac{x-5}{3}=\frac{y+2}{2}=\frac{z}{6}$ is $\qquad$ .

The Cartesian equation of a line is $2x - 3 = 3y + 1 = 5 - 6z$. The vector equation of the line passing through the point $(7, -5, 0)$ and parallel to the given line is