The equation of the plane containing the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to the plane containing the lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is

Let $P_1$ be the plane $3x - y - 7z = 11$ and $P_2$ be the plane passing through the points $(2, -1, 0)$,$(2, 0, -1)$,and $(5, 1, 1)$. If the foot of the perpendicular drawn from the point $(7, 4, -1)$ on the line of intersection of the planes $P_1$ and $P_2$ is $(\alpha, \beta, \gamma)$,then $\alpha + \beta + \gamma$ is equal to $............$.

If the distance of the point $(1, -2, 3)$ from the plane $x + 2y - 3z + 10 = 0$ measured parallel to the line $\frac{x-1}{3} = \frac{2-y}{m} = \frac{z+3}{1}$ is $\sqrt{\frac{7}{2}}$,then the value of $|m|$ is equal to ....... .

The equation of the plane passing through the point $(1,1,1)$ and through the line of intersection of $x+2y-z+1=0$ and $3x-y-4z+3=0$ is

If $d$ is the distance of the point $(2, 5, 10)$ from the plane containing the lines $\vec{r} = (4\hat{j} - \hat{k}) + \lambda(\hat{i} + 2\hat{j} - 2\hat{k})$ and $\vec{r} = (2\hat{i} + \hat{j}) + \mu(\hat{i} + 2\hat{j} - 2\hat{k})$, then $d^2 = $

Find the vector equation of the line passing through $(1, 2, 3)$ and perpendicular to the plane $\vec{r} \cdot (\hat{i} + 2\hat{j} - 5\hat{k}) + 9 = 0$.