Let a triangle PQR be such that P and Q lie on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ and are at a distance of 6 units from $R(1,2,3)$. If $(\alpha, \beta, \gamma)$ is the centroid of $\Delta P Q R$, then $\alpha+\beta+\gamma$ is equal to :

If the distance of the point $(a, 2,5)$ from the image of the point $(1,2,7)$ in the line $\frac{x}{1}=\frac{y-1}{1}=\frac{z-2}{2}$ is 4 , then the sum of all possible values of $a$ is equal to :

The square of the distance of the point $\mathrm{P}(5,6,7)$ from the line $\frac{x-2}{2}=\frac{y-5}{3}=\frac{z-2}{4}$ is equal to:

$\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(a \hat{i}-\hat{j}), a \neq 0$ and $\vec{r}=(4 \hat{i}-\hat{k})+\mu(2 \hat{i}+a \hat{k})$ from the origin is :