Let the equation of the plane,that passes through the point $(1,4,-3)$ and contains the line of intersection of the planes $3x-2y+4z-7=0$ and $x+5y-2z+9=0$,be $\alpha x+\beta y+\gamma z+3=0$. Then $\alpha+\beta+\gamma$ is equal to:

The plane through the intersection of planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to $Y$-axis also passes through the point

The foot of the perpendicular drawn from a point $A(1,1,1)$ onto a plane $\pi$ is $P(-3,3,5)$. If the equation of the plane parallel to the plane $\pi$ and passing through the midpoint of $AP$ is $ax-y+cz+d=0$,then $a+c-d=$

The point at which the line joining the points $(2, -3, 1)$ and $(3, -4, -5)$ intersects the plane $2x + y + z = 7$ is

$A$ plane $\pi_1$ contains the vectors $\bar{i}+\bar{j}$ and $\bar{i}+2\bar{j}$. Another plane $\pi_2$ contains the vectors $2\bar{i}-\bar{j}$ and $3\bar{i}+2\bar{k}$. $\bar{a}$ is a vector parallel to the line of intersection of $\pi_1$ and $\pi_2$. If the angle $\theta$ between $\bar{a}$ and $\bar{i}-2\bar{j}+2\bar{k}$ is acute, then $\theta=$

The equation of the plane passing through the line of intersection of planes $\pi_1: 2x + 6y + 4z - 7 = 0$ and $\pi_2: x - y - 2z - 2 = 0$, and perpendicular to the plane $x + y + 2z - 5 = 0$ is: