Let $$f(x) = x^{3}+ x^{2}f'(1)+2xf''(2)+f'''(3)$$, $$x\epsilon R$$. Then the value of f'(5) is :

We begin with $$f(x) = x^3 + x^2 f'(1) + 2xf''(2) + f'''(3)$$.

Writing $$f'(1) = a$$, $$f''(2) = b$$ and $$f'''(3) = c$$ yields $$f(x) = x^3 + ax^2 + 2bx + c$$.

Therefore $$f'(x) = 3x^2 + 2ax + 2b$$, $$f''(x) = 6x + 2a$$ and $$f'''(x) = 6$$.

Applying the third derivative condition gives $$f'''(3) = 6 = c$$.

Next, from the second derivative condition we have $$f''(2) = 12 + 2a = b$$.

Then the first derivative condition reads $$f'(1) = 3 + 2a + 2b = a$$.

From these, $$3 + 2a + 2b = a \Rightarrow a + 2b = -3$$ (i) and $$b = 12 + 2a$$ (ii).

Substituting (ii) into (i) yields $$a + 2(12 + 2a) = -3 \Rightarrow 5a + 24 = -3 \Rightarrow a = -\frac{27}{5}$$.

Substitution into (ii) then gives $$b = 12 + 2(-27/5) = 12 - 54/5 = 6/5$$.

Hence $$f'(x) = 3x^2 + 2(-27/5)x + 2(6/5) = 3x^2 - \frac{54}{5}x + \frac{12}{5}$$.

It follows that $$f'(5) = 75 - \frac{270}{5} + \frac{12}{5} = 75 - 54 + \frac{12}{5} = 21 + \frac{12}{5} = \frac{105 + 12}{5} = \frac{117}{5}$$.

The answer is Option 4: $$\frac{117}{5}$$.