Find the equation of the plane which passes through the points $(0,1,2)$ and $(-1,0,3)$ and is perpendicular to the plane $2x+3y+z=5$.

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 value of $k$ such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies on the plane $2x-4y+z=7$ is

The line of intersection of the planes $x + 2y = 0$ and $y - 3z + 3 = 0$ is

Let the plane $P: 4x - y + z = 10$ be rotated by an angle $\frac{\pi}{2}$ about its line of intersection with the plane $x + y - z = 4$. If $\alpha$ is the distance of the point $(2, 3, -4)$ from the new position of the plane $P$,then $35\alpha$ is

If the points $(1, 1, \lambda)$ and $(-3, 0, 1)$ are equidistant from the plane $3x + 4y - 12z + 13 = 0$,then the integer value of $\lambda$ is: