The point of intersection of the lines $\vec{r}=2 \vec{b}+t(6 \vec{c}-\vec{a})$ and $\vec{r}=\vec{a}+s(\vec{b}-3 \vec{c})$ is

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

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:

The length of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$ is (in $\text{ units}$)

Let $a, b \in R$. If the mirror image of the point $P(a, 6, 9)$ with respect to the line $\frac{x-3}{7} = \frac{y-2}{5} = \frac{z-1}{-9}$ is $(20, b, -a-9)$,then $|a+b|$ is equal to

If the shortest distance between the lines $\overline{r}_1 = \alpha \hat{i} + 2 \hat{j} + 2 \hat{k} + \lambda(\hat{i} - 2 \hat{j} + 2 \hat{k})$ and $\overline{r}_2 = -4 \hat{i} - \hat{k} + \mu(3 \hat{i} - 2 \hat{j} - 2 \hat{k})$ is $9$,where $\lambda, \mu \in R$ and $\alpha > 0$,then the value of $\alpha$ is: