$A$ ray of light passing through the point $A(1, 2, 3)$ strikes the plane $x+y+z=12$ at $B$ and on reflection passes through $C(3, 5, 9)$,then $OB$ is equal to

Let $A$ be a point having position vector $\bar{i}-3 \bar{j}$ and $\bar{r}=(\bar{i}-3 \bar{j})+t(\bar{j}-2 \bar{k})$ be a line. If $P$ is a point on this line and is at a minimum distance from the plane $\bar{r} \cdot(2 \bar{i}+3 \bar{j}+5 \bar{k})=0$, then the equation of the plane through $P$ and perpendicular to $AP$ is:

Let $P$ be a plane containing the line $\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+5}{2}$ and parallel to the line $\frac{x-3}{4}=\frac{y-2}{-3}=\frac{z+5}{7}$. If the point $(1, -1, \alpha)$ lies on the plane $P$,then the value of $|5\alpha|$ is equal to ....... .

If $P=(0,1,0)$ and $Q=(0,0,1)$,then the length of the projection of the line segment $PQ$ on the plane $x+y+z=3$ is:

The line $\frac{x - 1}{2} = -(y + 1) = \frac{z}{3}$ and the plane $3x + 2y - z = 5$ intersect at a point. The coordinates of the point are:

Find the equation of the plane containing the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ and the point $(0, 7, -7)$.