Official solution
f(x) = x3 + x2f '(1) + xf ''(2) + f '''(3)
\[ \Rightarrow \] f '(x) = 3x2 + 2xf '(1) + f ''(x) . . . . . (1)
\[ \Rightarrow \] f ''(x) = 6x + 2f '(1) . . . . . . (2)
\[ \Rightarrow \] 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)
\[ \Rightarrow \] 3f '(1) = - 15
\[ \Rightarrow \] f '(1) = - 5 \[ \Rightarrow \] f ''(2) = 2 . . . . .(2)
put x = 3 in equation (3) :
f ''' (3) = 6
\[ \therefore \] f(x) = x3 - 5x2 + 2x + 6
f(2) = 8 - 20 + 4 + 6 = - 2