To form the equation of a plane we need a point on it and normal vector $(\overrightarrow n)$

Step1

Given that the required plane contains the point $(1,-1,2)$

Let the normal to the required plane be $\overrightarrow n$

Also it is given that the required plane is $\perp$ to the planes

$2x+3y-2z-5=0$..............(i) and

$x+2y-3z-8$..........(ii)

Normal to the plane is $\overrightarrow n_1=(2,3,-2)$

Normal to the plane (ii) is $\overrightarrow n_2=(1,2,-3)$

Since the required plane is $\perp$ to (i) and (ii),

$\overrightarrow n=\overrightarrow n_1\times\overrightarrow n_2$

$\overrightarrow n=\left |\begin {array} \hat i & \hat j & \hat k\\2 & 3 & -2\\1 & 2 & -3\end {array}\right|=\hat i(-9+4)-\hat j(-6+2)+\hat k(4-3)$

$i.e., \overrightarrow n=(-5,4,1)$

Step 2

$\therefore$ Equation of the required plane is

$-5x+4y+z+d=0$

Given that this plane contains the point $(1,-1,2)$

$\Rightarrow\:$ It satisfies the equation of the plane.

$\Rightarrow\:-5\times 1+4\times (-1)+2+d=0$

$\Rightarrow\:d=7$

$\Rightarrow\:$ Equation of the required plane is $5x-4y-z-7=0$

Step 3

We know that distance of the point $(x_1,y_1,z_1)$ from the plane $ax+by+cz+d=0$ is given by

$\bigg|\large\frac{ax_1+by_1+cz_1+d}{\sqrt {a^2+b^2+c^2}}\bigg|$

The distance of the point $(-2,5,5)$ from the plane $5x-4y-z-7=0$ is given by

$\bigg|\large\frac{5\times (-2)+(-4)\times 5+(-1)\times 5-7}{\sqrt {5^2+(-4)^2+(-1)^2}}\bigg|$

$=\bigg|\large\frac{-10-20-5-7}{\sqrt 42}\bigg|$

$=\large\frac{42}{\sqrt {42}}$$=\sqrt {42}$ units.