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

Let $P$ be a plane passing through the points $(1,0,1), (1,-2,1)$ and $(0,1,-2)$. Let a vector $\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ be such that $\vec{a}$ is parallel to the plane $P$,perpendicular to $(\hat{i} + 2 \hat{j} + 3 \hat{k})$ and $\vec{a} \cdot (\hat{i} + \hat{j} + 2 \hat{k}) = 2$. Then $(\alpha - \beta + \gamma)^2$ equals:

Let $Q$ be the foot of the perpendicular from the point $P(7, -2, 13)$ on the plane containing the lines $\frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}$ and $\frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}$. Then $(PQ)^{2}$ is equal to ..... .

Find the equation of the plane which contains the line of intersection of the planes $\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4=0$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5=0$ and which is perpendicular to the plane $\vec{r} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})+8=0$.

If the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$,then $(\alpha, \beta)=$

The equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $3x+4y+5z=2$ and perpendicular to the $XY$-plane is