Thank you for reporting, we will resolve it shortly
Q.
The distance of the point $(1, -2, 4)$ from the plane passing through the point $(1, 2, 2)$ and perpendicular to the planes $x - y + 2z = 3$ and $2x - 2y + z + 12 = 0$, is :
Let equation of plane be
$a\left(x-1\right)+b\left(y-2\right)+c\left(z-2\right)=0$_____(1)
i) is perpendicular to given planes then
$a-b+2c=0$
$2a-2b+c=0$
Solving above equations $c = 0$ and $a = b$
equation of plane 1) can be
x + y - 3 = 0
distance from (1,-2,4) will be
$D=\frac{\left|1-2-3\right|}{\sqrt{1+1}}=\frac{4}{\sqrt{2}}=2\sqrt{2}$