Let $$f(x) = ax^3 + bx^2 + cx + 41$$ be such that $$f(1) = 40$$, $$f'(1) = 2$$ and $$f''(1) = 4$$. Then $$a^2 + b^2 + c^2$$ is equal to:

$$f(x) = ax^3 + bx^2 + cx + 41$$.

$$f(1) = a + b + c + 41 = 40 \implies a + b + c = -1$$ ... (i)

$$f'(x) = 3ax^2 + 2bx + c$$. $$f'(1) = 3a + 2b + c = 2$$ ... (ii)

$$f''(x) = 6ax + 2b$$. $$f''(1) = 6a + 2b = 4$$ ... (iii)

From (iii): $$3a + b = 2 \implies b = 2 - 3a$$.

From (ii): $$3a + 2(2-3a) + c = 2 \implies 3a + 4 - 6a + c = 2 \implies -3a + c = -2 \implies c = 3a - 2$$.

From (i): $$a + (2-3a) + (3a-2) = -1 \implies a = -1$$.

$$b = 2 - 3(-1) = 5$$, $$c = 3(-1) - 2 = -5$$.

$$a^2 + b^2 + c^2 = 1 + 25 + 25 = 51$$.

The correct answer is Option 3: 51.