Let P be the plane, which contains the line of intersection of the planes, x+y+z-6=0 and 2x+3y+z+ 5=0 and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to

Equation of plane $P$ which contains the line of intersection of planes $2 x + 3 y + z + 5 = 0$ and $X + y + Z - 6 = 0$ is

$2 x + 3 y + z + 5 + λ (x + y + z - 6) = 0$

$\Rightarrow (2 + λ) x + (3 + λ) y + (λ + 1) z + 5 - 6 λ = 0$ ..(i)

Since, plane $P$ is .. $L$ to $x y$ -plane

$∴ λ + 1 = 0 \Rightarrow λ = - 1$

$∴$ from (i), equation of plane $P$ is

$x + 2 y + 11 = 0$

Now, distance of point

$A (0, 0, 256)$ from plan $P$ is

$\frac{| 0 (1) + 0 (2) + 256 (0) + 11 |}{\sqrt{1^{2} + 2^{2}}} = \frac{11}{\sqrt{5}}$

Let $P$ be the plane, which contains the line of intersection of the planes, $x + y + z - 6 = 0$ and $2 x + 3 y + z + 5 = 0$ and it is perpendicular to the xy-plane. Then the distance of the point $(0, 0, 256)$ from P is equal to