Let f:R→R be a function such that f(x)=x³+x²f'(1)+xf''(2)+f'''(3), x∈R. Then f(2) equals?

f(x)=x³+x²f'(1)+xf''(2)+f'''(3) ⇒f'(x)=3x²+2xf'(1)+f''(2) (1) ⇒f''(x)=6x+2f'(1). (2) ⇒f'''(x)=6 (3)
Put x=1 in equation(1): f'(1)=3+2f'(1)+f''(2). (4)
Put x=2 in equation(2): f''(2)=12+2f'(1) (5)
From equation (4) (5): 3-f'(1)=12+2f'(1) ⇒3f'(1)=-9 ⇒f'(1)=-3 ⇒f''(2)=6. (2)
Put x=3 in equation (3): f'''(3)=6
∴f(x)=x³-3x²+6x+6
f(2)=8-12+12+6=14