Step 1: Understanding the Concept:
To construct the equation of a plane, we require a point on it and its normal vector. Since the required plane is perpendicular to two given planes, its normal vector will be perpendicular to the normal vectors of both given planes. The cross product of the two given normals will yield the normal of our required plane.
Step 2: Key Formula or Approach:
1. Normal vector $\bar{n} = \bar{n}_1 \times \bar{n}_2$, where $\bar{n}_1, \bar{n}_2$ are normals of the given planes.
2. Equation of a plane through $(x_1, y_1, z_1)$ with normal $\langle a, b, c \rangle$ is $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.
3. Distance of point $(x_0, y_0, z_0)$ from plane $ax+by+cz+d=0$ is $d = \frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}$.
Step 3: Detailed Explanation:
The normal vectors of the two given planes are $\bar{n}_1 = \langle 2, -2, 1 \rangle$ and $\bar{n}_2 = \langle 1, -1, 2 \rangle$.
Compute the cross product to get the normal vector $\bar{n}$ of the required plane:
\[ \bar{n} = \bar{n}_1 \times \bar{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & -2 & 1
1 & -1 & 2 \end{vmatrix} \]
\[ \bar{n} = \hat{i}(-4 - (-1)) - \hat{j}(4 - 1) + \hat{k}(-2 - (-2)) \]
\[ \bar{n} = -3\hat{i} - 3\hat{j} + 0\hat{k} = \langle -3, -3, 0 \rangle \]
To simplify the equation, we can scale this normal vector by dividing by $-3$ to get a proportional normal vector: $\bar{n}' = \langle 1, 1, 0 \rangle$.
Now, form the equation of the plane passing through the point $(1, -2, 1)$ using $\bar{n}'$:
\[ 1(x - 1) + 1(y - (-2)) + 0(z - 1) = 0 \]
\[ x - 1 + y + 2 = 0 \implies x + y + 1 = 0 \]
We now find the perpendicular distance of the target point $(1, 2, 2)$ from this plane $x + y + 1 = 0$:
\[ \text{Distance} = \frac{|1(1) + 1(2) + 0(2) + 1|}{\sqrt{1^2 + 1^2 + 0^2}} \]
\[ \text{Distance} = \frac{|1 + 2 + 1|}{\sqrt{2}} = \frac{4}{\sqrt{2}} \]
Rationalize the denominator:
\[ \text{Distance} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \]
Step 4: Final Answer:
The distance is $2\sqrt{2}$ units.