The foot of the perpendicular drawn from the point $(4,2,3)$ to the line joining the points $(1,-2,3)$ and $(1,1,0)$ lies on the plane

If the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$ makes an angle $\alpha$ with the positive $x$-axis,then $\cos \alpha = $

Statement-$1$: The point $A(1, 0, 7)$ is the reflection of the point $B(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$.
Statement-$2$: The line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$ is the perpendicular bisector of the line segment joining $A(1, 0, 7)$ and $B(1, 6, 3)$.

If the line $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$ lies in the plane $2x - 4y + z = 7$,then $k = \dots$

Let a line $l$ pass through the origin and be perpendicular to the lines $l_1: \overrightarrow{r} = (\hat{i} - 11\hat{j} - 7\hat{k}) + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$ and $l_2: \overrightarrow{r} = (-\hat{i} + \hat{k}) + \mu(2\hat{i} + 2\hat{j} + \hat{k})$. If $P$ is the point of intersection of $l$ and $l_1$,and $Q(\alpha, \beta, \gamma)$ is the foot of the perpendicular from $P$ on $l_2$,then $9(\alpha + \beta + \gamma)$ is equal to:

The equation of the plane passing through the intersection of the planes $x + 2y + 3z + 4 = 0$ and $4x + 3y + 2z + 1 = 0$ and the origin is