Let $P$ be the foot of the perpendicular from the point $Q(10,-3,-1)$ on the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}$. Then the area of the right-angled triangle $PQR$,where $R$ is the point $(3,-2,1)$,is

The shortest distance between lines $\overline{r}=(2 \hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$ and $\overline{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+\hat{j}-5 \hat{k})$ is

Let the image of the point $P(1, 2, 3)$ in the line $L : \frac{x-6}{3} = \frac{y-1}{2} = \frac{z-2}{3}$ be $Q$. Let $R(\alpha, \beta, \gamma)$ be a point that divides the line segment $PQ$ internally in the ratio $1:3$. Then the value of $22(\alpha+\beta+\gamma)$ is equal to

The lines $\frac{x - 1}{2} = \frac{y - 1}{2} = \frac{z - 3}{0}$ and $\frac{x - 2}{0} = \frac{y - 3}{0} = \frac{z - 4}{1}$ are:

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$(s)$ $L_3$?
$(1) \overrightarrow{r} = \frac{1}{3}(2\hat{i} + \hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}), t \in R$
$(2) \overrightarrow{r} = \frac{2}{9}(2\hat{i} - \hat{j} + 2\hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}), t \in R$
$(3) \overrightarrow{r} = t(2\hat{i} + 2\hat{j} - \hat{k}), t \in R$
$(4) \overrightarrow{r} = \frac{2}{9}(4\hat{i} + \hat{j} + \hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}), t \in R$

The coordinates of the point of intersection of the lines $\frac{x - 3}{1} = \frac{y - 5}{2} = \frac{z - 1}{-1}$ and $\frac{x - 4}{2} = \frac{y - 2}{-1} = \frac{z - 4}{2}$ are: