$A$ line drawn from the point $A(1, 3, 2)$ parallel to the line $\frac{x}{2} = \frac{y}{4} = \frac{z}{1}$ intersects the plane $3x + y + 2z = 5$ at point $B$. Then the coordinates of point $B$ are:

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

In $R^3$,consider the planes $P_1: y=0$ and $P_2: x+z=1$. Let $P_3$ be a plane,different from $P_1$ and $P_2$,which passes through the intersection of $P_1$ and $P_2$. If the distance of the point $(0,1,0)$ from $P_3$ is $1$ and the distance of a point $(\alpha, \beta, \gamma)$ from $P_3$ is $2$,then which of the following relations is (are) true?
$(A)$ $2\alpha+\beta+2\gamma+2=0$
$(B)$ $2\alpha-\beta+2\gamma+4=0$
$(C)$ $2\alpha+\beta-2\gamma-10=0$
$(D)$ $2\alpha-\beta+2\gamma-8=0$

Let the line $L$ pass through the point $(0,1,2)$,intersect the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and be parallel to the plane $2x+y-3z=4$. Then the distance of the point $P(1,-9,2)$ from the line $L$ is

The $XOZ$ plane divides the line segment joining the points $(2, 3, 1)$ and $(6, 7, 1)$ in the ratio: (in $:7$)

The equation of the plane passing through the point of intersection of the planes $2x-y+z-3=0$ and $4x-3y+5z+9=0$ and parallel to the line $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z-3}{5}$ is $\alpha x+\beta y+\gamma z+d=0$. Then $\alpha+\beta+\gamma+d=$