If lines $\frac{x-3}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{6-z}{5}$ are perpendicular to each other,then $k=$ $\qquad$ .

When are the two lines $x = ay + b, z = cy + d$ and $x = a'y + b', z = c'y + d'$ perpendicular to each other?

Let a line $L$ passing through the point $(1, 1, 1)$ be perpendicular to both the vectors $2\hat{i} + 2\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 2\hat{k}$. If $P(a, b, c)$ is the foot of the perpendicular from the origin on the line $L$, then the value of $34(a + b + c)$ is:

If the Cartesian equation of the line is $x-1=2y+3=3-z$,then its vector equation is

Let $A(2,3,-1), B(4,1,0), C(-1,-1,11)$ be the vertices of a triangle $ABC$. Let $D$ be the point where the bisector of $\angle BAC$ meets the side $BC$. Then the direction ratios of $AD$ are

The angle between the lines $\vec{r}=(2 \hat{i}-3 \hat{j}+\hat{k})+\lambda(\hat{i}+4 \hat{j}+3 \hat{k})$ and $\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}+2 \hat{j}-3 \hat{k})$ is