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

If the shortest distance between the lines $\frac{x - 1}{\alpha} = \frac{y + 1}{-1} = \frac{z}{1}, (\alpha \ne -1)$ and $x + y + z + 1 = 0 = 2x - y + z + 3$ is $\frac{1}{\sqrt{3}}$,then a value of $\alpha$ is

The acute angle between the line $\bar{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})$ and the plane $\bar{r} \cdot(2\hat{i}-\hat{j}+\hat{k})=5$ is

The equation of the line passing through the point $(2, 3, 1)$ and parallel to the line of intersection of the planes $x - 2y - z + 5 = 0$ and $x + y + 3z = 6$ is:

If the distance of the point $2 \hat{i} + 3 \hat{j} + \lambda \hat{k}$ from the plane $\vec{r} \cdot (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) = 13$ is $5$ units,then $\lambda =$

Consider the planes $3x - 6y - 2z = 15$ and $2x + y - 2z = 5$.
$STATEMENT-1$ : The parametric equations of the line of intersection of the given planes are $x = 3 + 14t, y = 1 + 2t, z = 15t$ because
$STATEMENT-2$ : The vector $14\hat{i} + 2\hat{j} + 15\hat{k}$ is parallel to the line of intersection of the given planes.