The shortest distance between the lines $\frac{x+7}{-6}=\frac{y-6}{7}=\frac{z}{1}$ and $\frac{7-x}{2}=y-2=z-6$ is

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

Prove that the line through $A(0,-1,-1)$ and $B(4,5,1)$ intersects the line through $C(3,9,4)$ and $D(-4,4,4)$.

The lines $\frac{6x-6}{18} = \frac{y+1}{3} = \frac{z-1}{5}$ and $\frac{3x+6}{12} = \frac{y-1}{3} = \frac{z+1}{2}$ are $\dots$

If $A(1,2,3), B(3,7,-2), C(6,7,7)$ and $D(-1,0,-1)$ are points in a plane,then the vector equation of the line passing through the centroids of $\triangle ABD$ and $\triangle ACD$ is

If the shortest distance between the lines $\vec{r}_{1}=\alpha \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k}), \lambda \in R, \alpha>0$ and $\vec{r}_{2}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k}), \mu \in R$ is $9$,then $\alpha$ is equal to $.....$