To solve the equation \( 9x^2 + y^2 + 2z^2 - 18x - 6y + 3z + 20 = 0 \), we can complete the square for each variable.

### Step 1: Complete the square for \( x \)

For \( 9x^2 - 18x \):
\[
9(x^2 - 2x) = 9((x - 1)^2 - 1) = 9(x - 1)^2 - 9
\]

### Step 2: Complete the square for \( y \)

For \( y^2 - 6y \):
\[
(y^2 - 6y) = (y - 3)^2 - 9
\]

### Step 3: Complete the square for \( z \)

For \( 2z^2 + 3z \):
\[
2(z^2 + \frac{3}{2}z) = 2((z + \frac{3}{4})^2 - \frac{9}{16}) = 2(z + \frac{3}{4})^2 - \frac{9}{8}
\]

### Step 4: Substitute back into the equation

We have:
\[
9(x - 1)^2 - 9 + (y - 3)^2 - 9 + 2(z + \frac{3}{4})^2 - \frac{9}{8} + 20 = 0
\]
Combine the constants:
\[
9(x - 1)^2 + (y - 3)^2 + 2(z + \frac{3}{4})^2 + 20 - 9 - 9 - \frac{9}{8} = 0
\]
\[
9(x - 1)^2 + (y - 3)^2 + 2(z + \frac{3}{4})^2 + \frac{160}{8} - \frac{18}{8} - \frac{9}{8} = 0
\]
\[
9(x - 1)^2 + (y - 3)^2 + 2(z + \frac{3}{4})^2 + \frac{133}{8} = 0
\]

Since all squares are non-negative, the left side can equal zero only if each square is zero.

### Step 5: Set each term to zero

1. \( 9(x - 1)^2 = 0 \) → \( x = 1 \)
2. \( (y - 3)^2 = 0 \) → \( y = 3 \)
3. \( 2(z + \frac{3}{4})^2 = 0 \) → \( z = -\frac{3}{4} \)

### Step 6: Calculate \( x^{2022} + y - 3z^{10} \)

Now substitute \( x = 1 \), \( y = 3 \), \( z = -\frac{3}{4} \):

1. \( x^{2022} = 1^{2022} = 1 \)
2. \( y = 3 \)
3. \( z^{10} = \left(-\frac{3}{4}\right)^{10} = \left(\frac{3}{4}\right)^{10} \)

Calculating \( \left(\frac{3}{4}\right)^{10} \):
\[
\left(\frac{3}{4}\right)^{10} = \frac{3^{10}}{4^{10}} = \frac{59049}{1048576}
\]

Now compute:
\[
x^{2022} + y - 3z^{10} = 1 + 3 - 3 \cdot \frac{59049}{1048576}
\]
\[
= 4 - \frac{177147}{1048576}
\]
\[
= \frac{4194304 - 177147}{1048576} = \frac{4017157}{1048576}
\]

Thus, the final answer is:
\[
\boxed{\frac{4017157}{1048576}}
\]