Let the number $$(22)^{2022} + (2022)^{22}$$ leave the remainder $$\alpha$$ when divided by 3 and $$\beta$$ when divided by 7. Then $$(\alpha^2 + \beta^2)$$ is equal to

$$22^{2022} + 2022^{22} \pmod{3}$$:

$$22 \equiv 1 \pmod{3}$$, so $$22^{2022} \equiv 1$$. $$2022 \equiv 0 \pmod{3}$$, so $$2022^{22} \equiv 0$$.

$$\alpha = 1 + 0 = 1$$.

$$22^{2022} + 2022^{22} \pmod{7}$$:

$$22 \equiv 1 \pmod{7}$$, so $$22^{2022} \equiv 1$$. $$2022 = 288 \times 7 + 6$$, so $$2022 \equiv -1 \pmod{7}$$, $$2022^{22} \equiv 1$$.

$$\beta = 1 + 1 = 2$$.

$$\alpha^2 + \beta^2 = 1 + 4 = 5$$.

The correct answer is Option 3: 5.