If $A(1,1,2)$,$B(4,2,1)$ and $C(2,3,5)$ are the vertices of a triangle,then a vector representing the median of the triangle through $A$ 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 lines $\frac{x - 1}{k} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x - 2}{3} = \frac{y - 3}{k} = \frac{z - 1}{2}$ intersect,find the value of $k$.

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3}$ is

Consider the lines $L_1: x-1=y-2=z$ and $L_2: x-2=y=z-1$. Let the feet of the perpendiculars from the point $P(5,1,-3)$ on the lines $L_1$ and $L_2$ be $Q$ and $R$ respectively. If the area of the triangle $PQR$ is $A$,then $4A^2$ is equal to:

If the line $\frac{2-x}{3}=\frac{3y-2}{4\lambda+1}=4-z$ makes a right angle with the line $\frac{x+3}{3\mu}=\frac{1-2y}{6}=\frac{5-z}{7}$,then $4\lambda+9\mu$ is equal to :