If the mirror image of the point $P(3, 4, 9)$ in the line $\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$,then $14(\alpha+\beta+\gamma)$ is :

Let $L_1$ and $L_2$ denote the lines $\overrightarrow{r} = \hat{i} + \lambda(-\hat{i} + 2\hat{j} + 2\hat{k}), \lambda \in R$ and $\overrightarrow{r} = \mu(2\hat{i} - \hat{j} + 2\hat{k}), \mu \in R$ respectively. If $L_3$ is a line which is perpendicular to both $L_1$ and $L_2$ and intersects both of them,then which of the following options describe$(s)$ $L_3$?
$(1) \overrightarrow{r} = \frac{1}{3}(2\hat{i} + \hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}), t \in R$
$(2) \overrightarrow{r} = \frac{2}{9}(2\hat{i} - \hat{j} + 2\hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}), t \in R$
$(3) \overrightarrow{r} = t(2\hat{i} + 2\hat{j} - \hat{k}), t \in R$
$(4) \overrightarrow{r} = \frac{2}{9}(4\hat{i} + \hat{j} + \hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}), t \in R$

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$,and the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0}$ is $\alpha$,then $28 \alpha^2$ is equal to $........$.

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

Find the coordinates of the foot of the perpendicular drawn from the point $(1, 0, 0)$ to the line $\frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z + 10}{8}$.

Find the vector equation of the line which is parallel to the vector $3 \hat{i}-2 \hat{j}+6 \hat{k}$ and which passes through the point $(1, -2, 3)$.