Let the foot of the perpendicular from the point $A(4, 3, 1)$ on the plane $P: x - y + 2z + 3 = 0$ be $N$. If $B(5, \alpha, \beta)$,where $\alpha, \beta \in \mathbb{Z}$,is a point on the plane $P$ such that the area of the triangle $ABN$ is $3\sqrt{2}$,then $\alpha^2 + \beta^2 + \alpha\beta$ is equal to $...........$.

Find the equation of the line passing through the point $(3, 0, 1)$ and parallel to the planes $x+2y=0$ and $3y-z=0$.

$A$ line with direction ratios $1, -1, 2$ intersects the lines $\frac{x}{2} = \frac{y}{3} = \frac{z+1}{3}$ and $\frac{x+1}{-1} = \frac{y-2}{1} = \frac{z}{4}$ at the points $P$ and $Q$, respectively. If the length of the line segment $PQ$ is $\alpha$, then $225\alpha^2$ is equal to:

Let two vertices of triangle $ABC$ be $(2,4,6)$ and $(0,-2,-5)$,and its centroid be $(2,1,-1)$. If the image of the third vertex in the plane $x+2y+4z=11$ is $(\alpha, \beta, \gamma)$,then $\alpha \beta+\beta \gamma+\gamma \alpha$ is equal to

Let the line $L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1}$ intersect the plane $2x+y+3z=16$ at the point $P$. Let the point $Q$ be the foot of the perpendicular from the point $R(1, -1, -3)$ on the line $L$. If $\alpha$ is the area of triangle $PQR$,then $\alpha^2$ is equal to $...........$.

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 $P_1 + \lambda P_2 = 0, \lambda \in \mathbb{R}$.