If a line $L$ is the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If line $L$ makes an angle $\alpha$ with the positive $X$-axis,then the value of $\sec \alpha$ is

The vector equation of the plane containing the lines $r = (i + j) + \lambda (i + 2j - k)$ and $r = (i + j) + \mu (-i + j - 2k)$ is

$A$ plane which passes through the point $(3, 2, 0)$ and the line $\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}$ is

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

The vector equation of the plane passing through the line of intersection of the planes $x + y + z = 1$ and $2x + 3y + 4z = 5$ which is perpendicular to the plane $x - y + z = 0$ is

Find the position vector of the point where the line $r = (i - j + k) + t(i + j + k)$ meets the plane $r \cdot (i + j - k) = 5$.