Let two vertices of triangle $ABC$ be $(2,4,6)$ and $(0,-2,-5)$,and its centroid be $(2,1,-1)$. If the image of the third vertex in the plane $x+2y+4z=11$ is $(\alpha, \beta, \gamma)$,then $\alpha \beta+\beta \gamma+\gamma \alpha$ is equal to

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

What is the length of the perpendicular drawn from the origin to the plane passing through point $a$ and containing the line $r = b + \lambda c$?

If the line $\frac{x-3}{2}=\frac{y+5}{-1}=\frac{z+2}{2}$ lies in the plane $\alpha x+3y-z+\beta=0$,then the values of $\alpha$ and $\beta$ respectively are ....

If the equation of the line passing through the point $(0, -\frac{1}{2}, 0)$ and perpendicular to the lines $\overrightarrow{r} = \lambda(\hat{i} + a\hat{j} + b\hat{k})$ and $\overrightarrow{r} = (\hat{i} - \hat{j} - 6\hat{k}) + \mu(-b\hat{i} + a\hat{j} + 5\hat{k})$ is $\frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4}$,then $a+b+c+d$ is equal to :

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