Internal discussion!

We have $f(x)=x2+xg′(1)+g′′(2)$

Scince these are constants, let

$\begin{aligned} g'(1) &= m \\ g''(2) &= n \\ f(1) &= o \end{aligned}$

Let's beging with $f(x)$

$\begin{aligned} f(x) &= x^2 + mx + n \\ \Rightarrow f'(x) &= 2x + m\\ \text{And, } f''(x) &= 2 \end{aligned}$

Now,

$\begin{aligned} g(x) &= f(1)x^2 + xf'(x) + f''(x)\\ &=ox^2 + 2x^2 + mx + 2 \\ \Rightarrow g'(x) &= 2ox + 4x +m\\ \Rightarrow g'(1) = m &= 2o + 4 + m\\ \Rightarrow o &= -2\\ &\boxed{f(1) = -2}\\\\ \Rightarrow g'(x) &= -4x + 4x + g'(1)\\ &=g'(1)\\ \text{And, } g''(x) &= 0\\ \Rightarrow n &= 0\\ &\boxed{g''(2) = 0}\\\\ \text{Now, } f(1) = -2 &= 1 +m + 0\\ \Rightarrow a &= -3\\ &\boxed{g'(1) = -3}\\\\ \text{Hence } f(x) &= x^2 - 3x = ax^2 + bx\\ g'(x) &= -3 = -c\\ a + 2b+3c& = 1-6+9 = 4\\\\ &\boxed{\therefore a +2b + 3c = 4} \end{aligned}$

Verifying,

$\begin{aligned}f(x) &= x^2 - 3x\\ f(1) &= -2\\ f'(x) &= 2x - 3\\ g(x) &= -2x^2 + 2x^2 -3x + 2\\ &=-3x + 2\\ \Rightarrow g'(x) &= -3\\ \Rightarrow g''(x) &= 0\\\\ f(x)&=x^2+xg′(1)+g′′(2) \\ &= x^2 - 3x\end{aligned}$