The line $\frac{x + 3}{3} = \frac{y - 2}{-2} = \frac{z + 1}{1}$ and the plane $4x + 5y + 3z - 5 = 0$ intersect at a point

$L$ is a line passing through the point $A(1, 0, -3)$ and parallel to a line having direction ratios $0, 1, -2$. $P$ is a point on the line $L$ which is at a minimum distance from the plane $2x + 3y + 5z = 1$. Then,the equation of the plane through $P$ and perpendicular to $AP$ is

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

The line of intersection of the planes $\vec r \cdot (3\hat i - \hat j + \hat k) = 1$ and $\vec r \cdot (\hat i + 4\hat j - 2\hat k) = 2$ is:

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

Let $P$ be the plane passing through the line $\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$ and the point $(2,4,-3)$. If the image of the point $(-1,3,4)$ in the plane $P$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to