Let the line $\ell: x = \frac{1-y}{-2} = \frac{z-3}{\lambda}, \lambda \in R$ meet the plane $P: x + 2y + 3z = 4$ at the point $(\alpha, \beta, \gamma)$. If the angle between the line $\ell$ and the plane $P$ is $\cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$,then $\alpha + 2\beta + 6\gamma$ is equal to

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 the value of $\alpha \beta$ is:

Let the image of the point $\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$ in the plane $x-2y+z-2=0$ be $P$. If the distance of the point $Q(6, -2, \alpha)$,where $\alpha > 0$,from $P$ is $13$,then $\alpha$ is equal to $...........$.

The distance of the point $(1, 6, 2)$ from the point of intersection of the line $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$ and the plane $x-y+z=16$ is (in $\text{ units}$)

The distance of the point $(-1, -5, -10)$ from the point of intersection of the line $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$ and the plane $x-y+z=5$ is

Assertion $(A)$: The equation of the plane passing through the point $(4, 4, 4)$ and the intersection of the planes $x + y + z = 6$ and $2x + 3y + 4z = 0$ is $29x + 23y + 17z = 276$.
Reason $(R)$: The equation of the plane passing through the line of intersection of planes $P_1 = 0$ and $P_2 = 0$ is $P_1 + \lambda P_2 = 0, \lambda \in \mathbb{R}$.