Let  $f\left(x\right)=a{x}^2+bx+c$ be such that   $f\left(1\right)=3,f\left(-2\right)=\lambda$ and  $f\left(3\right)=4$. If  $f\left(0\right)+f\left(1\right)+f\left(-2\right)+f\left(3\right)=14$  then  $\lambda$ is

$f\left(1\right)=a+b+c=3\_\_\_\_\_\left(1\right)$

$f\left(-2\right)=4a-2b+c=\lambda \_\_\_\_\_\left(2\right)$

$f\left(3\right)=9a+3b+c=4\_\_\_\_\_\left(3\right)$

$f\left(0\right)=c\_\_\_\_\_\left(4\right)$

From (2) & (3)  $\Rightarrow a+b=\frac{4-\lambda }{5}$

Also   $c+3+\lambda +4=14\Rightarrow c=7-\lambda$

From $\left(1\right)\Rightarrow \frac{4-\lambda }{5}+7-\lambda =3$

$\Rightarrow \lambda =4$