 Consider non-negative real numbers $a, b, c$ satisfying the condition $a^2 + b^2 + c^2 = 2$ . Find the maximum value of the following expression $$P=\frac{\sqrt{b^2+c^2}}{3-a}+\frac{\sqrt{c^2+a^2}}{3-b}+a+b-2022c$$
1. Given the non-negative real numbers \(a, b, c\) satisfying \(a^2 + b^2 + c^2 = 2\), we need to find the maximum value of the expression:
   \[
   P = \frac{\sqrt{b^2+c^2}}{3-a} + \frac{\sqrt{c^2+a^2}}{3-b} + a + b - 2022c
   \]

2. We start by considering the cyclic sum:
   \[
   P = \sum_{\text{cyc}} \left( \frac{\sqrt{2-a^2}}{3-a} + a - \frac{3}{8}(a^2-1) - \frac{3}{2} \right) + 3 + \frac{3}{8}(a^2 + b^2 - 2) - 2022c
   \]

3. Simplifying the expression inside the sum:
   \[
   \frac{\sqrt{2-a^2}}{3-a} + a - \frac{3}{8}(a^2-1) - \frac{3}{2} \leq 0
   \]

4. To prove the above inequality, we need to show:
   \[
   8\sqrt{2-a^2} \leq (3-a)(3a^2 - 8a + 9)
   \]

5. Squaring both sides to eliminate the square root:
   \[
   64(2-a^2) \leq (3-a)^2(3a^2 - 8a + 9)^2
   \]

6. Simplifying the left-hand side:
   \[
   128 - 64a^2
   \]

7. Expanding the right-hand side:
   \[
   (3-a)^2(3a^2 - 8a + 9)^2 = (9 - 6a + a^2)(9a^4 - 48a^3 + 78a^2 - 48a + 9)
   \]

8. After expanding and simplifying, we get:
   \[
   (a-1)^2(9a^4 - 84a^3 + 310a^2 - 580a + 601) \geq 0
   \]

9. This inequality holds because:
   \[
   9a^4 - 84a^3 + 310a^2 - 580a + 601 = (3a^2 - 14a + 6)^2 + 78a^2 - 412a + 565 > 0
   \]

10. Therefore, the inequality is true, and we have:
    \[
    P \leq 3 + \frac{3}{8}(a^2 + b^2 + c^2 - 2) = 3
    \]

11. The equality occurs when \(a = b = 1\) and \(c = 0\).

The final answer is \(\boxed{3}\).