The line passing through the points $(a, 1, 6)$ and $(3, 4, b)$ crosses the $yz$-plane at $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$. Then the value of $(3a + 4b)$ is:

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 points $A(3, 2, 0)$,$B(5, 3, 2)$,and $C(-9, 6, -3)$ are the vertices of a triangle $ABC$. If $AD$ is the internal bisector of $\angle BAC$ which meets $BC$ at $D$,then the coordinates of $D$ are:

The sum of all values of $ \alpha $, for which the shortest distance between the lines $ \frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha} $ and $ \frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2\alpha} $ is $ \sqrt{2} $, is

The shortest distance between the lines $\vec{r} = (\frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}) + \lambda(2\hat{i} - 5\hat{j} + 6\hat{k})$ and $\vec{r} = (-\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k}) + \mu(\hat{i} - \hat{k})$, where $\lambda, \mu \in R$, is:

Find the shortest distance between the lines $\vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=(2 \hat{i}-\hat{j}-\hat{k})+\mu(2 \hat{i}+\hat{j}+2 \hat{k})$.