If $f(1)=f(2)=0$ then exist $c$ such that $cf''(c)+2f'(c)=0$

Let the function $f$ be continuous on $[0,2]$, second-order differentiable on the interval $(0,2)$ and satisfy the condition $f(1)=f(2)=0$. Prove that there exists a real number $c$ in $(0,2)$ such that $cf''(c)+2f'(c)=0$

My idea is use MVT to prove that $f'(c)=\frac{-f(0)}{2}$ and use idea from problem If f(0)=f(1)=f(2)=0, ∀x,∃c,f(x)=16x(x−1)(x−2)f′′′(c) but I stuck to prove it. Can someone help me please, thank you.