Find the angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane $10x+2y-11z=3$.

$\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})=5$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})=3$ are two planes. $A$ plane $\pi$ passing through the line of intersection of these two planes, passes through the point $(0,1,2)$. If the equation of $\pi$ is $\vec{r} \cdot(a \hat{i}+b \hat{j}+c \hat{k})=m$, then $\frac{b c}{a^2}=$

The distance between the line $\vec{r} = (2\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + 4\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$ is:

The ratio in which the plane $x - 2y + 3z = 17$ divides the line segment joining the points $A(-2, 4, 7)$ and $B(3, -5, 8)$ is:

The distance of the point $(-1, 2, 3)$ from the plane $\vec{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 10$ measured parallel to the line of the shortest distance between the lines $\vec{r} = (\hat{i} - \hat{j}) + \lambda(2\hat{i} + \hat{k})$ and $\vec{r} = (2\hat{i} - \hat{j}) + \mu(\hat{i} - \hat{j} + \hat{k})$ is:

If $L$ is the line of intersection of two planes $x+2y+2z=15$ and $x-y+z=4$ and the direction ratios of the line $L$ are $(a, b, c)$, then $\frac{a^2+b^2+c^2}{b^2}=$