Given $f\left(x\right)=a{x}^2+bx+c$ 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 equal to

Now given $f\left(0\right)+f\left(1\right)+f\left(-2\right)+f\left(3\right)=14$

$\Rightarrow f\left(0\right)+3+\lambda +4=14$

$\Rightarrow f\left(0\right)=7-\lambda =c$

Now $f\left(1\right)=a+b+c=3 ...\left(\mathrm{i}\right)$

$f\left(3\right)=9a+3b+c=4 ...\left(\mathrm{ii}\right)$

$f\left(-2\right)=4a-2b+c=\lambda ...\left(\mathrm{iii}\right)$

Now subtracting equation $\left(\mathrm{ii}\right)-\left(\mathrm{iii}\right)$

We get, $a+b=\frac{4-\lambda }{5}$, now putting  in equation $\left(\mathrm{i}\right)$

We get, $\frac{4-\lambda }{5}+7-\lambda =3$

$\Rightarrow 6\lambda =24$

$\Rightarrow \lambda =4$