Let, $$f: R \to R$$ be a function such that $$f(x) = x^3 + x^2f'(1) + xf''(2) + f'''(3)$$, $$\forall x \in R$$. Then $$f(2)$$ equals:

We are given that the function $$f:\mathbb R\to\mathbb R$$ satisfies, for every real $$x,$$

$$f(x)=x^{3}+x^{2}f'(1)+x\,f''(2)+f'''(3).$$

For convenience, let us denote the three unknown constants that appear on the right-hand side as

$$A=f'(1),\qquad B=f''(2),\qquad C=f'''(3).$$

With this notation the functional equation becomes

$$f(x)=x^{3}+Ax^{2}+Bx+C\qquad\forall x\in\mathbb R.$$

Thus $$f(x)$$ is itself the cubic polynomial

$$f(x)=x^{3}+Ax^{2}+Bx+C.$$

Now we differentiate this expression successively, always using the basic power-rule $$\dfrac{d}{dx}(x^{n})=nx^{n-1}.$$

First derivative:

$$f'(x)=3x^{2}+2Ax+B.$$

Second derivative:

$$f''(x)=6x+2A.$$

Third derivative:

$$f'''(x)=6.$$

Notice that $$f'''(x)=6$$ is a constant, independent of $$x.$$ Evaluating this at $$x=3$$ gives

$$C=f'''(3)=6.$$

Next, we evaluate $$f''(x)$$ at $$x=2$$ in order to relate $$B$$ and $$A$$($$\text{since }$$B=f''(2)).$$

$$f''(2)=6\cdot 2+2A=12+2A.$$

So we have

$$B=12+2A.$$

Finally, we evaluate $$f'(x)$$ at $$x=1$$ because $$A=f'(1).$$

$$f'(1)=3(1)^{2}+2A(1)+B=3+2A+B.$$

But by definition $$f'(1)=A,$$ hence

$$A=3+2A+B.$$

We now solve the two simultaneous equations

$$\begin{cases} B=12+2A,\\ A=3+2A+B. \end{cases}$$

Substituting $$B=12+2A$$ into the second equation gives

$$A=3+2A+(12+2A)=3+12+4A.$$

$$A=15+4A.$$

Bringing all terms to one side,

$$A-4A=15\;\Longrightarrow\;-3A=15\;\Longrightarrow\;A=-5.$$

Using $$A=-5$$ in $$B=12+2A$$ yields

$$B=12+2(-5)=12-10=2.$$

We already have $$C=6.$$

Therefore the explicit form of the function is

$$f(x)=x^{3}-5x^{2}+2x+6.$$

We are asked to find $$f(2).$$ Substituting $$x=2$$ in the polynomial,

$$\begin{aligned} f(2)&=(2)^{3}-5(2)^{2}+2(2)+6\\ &=8-5\cdot 4+4+6\\ &=8-20+4+6\\ &=18-20\\ &=-2. \end{aligned}$$

Hence, the correct answer is Option D.