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 equal to

 

$$\begin{gathered} f\left(0\right)+3+\lambda +4=14 \\ \therefore f\left(0\right)=7-\lambda =c \\ f\left(1\right)=a+b+c=3 ...\left(i\right) \\ f\left(3\right)=9a+3b+c=4 .....\left(ii\right) \\ f(-2)=4a-2b+c=\lambda ......\left(iii\right) \\ \left(ii\right)-\left(iii\right)\Rightarrow a+b=\frac{4-\lambda }{5} put in equation \left(i\right) \\ \Rightarrow \frac{4-\lambda }{5}+7-\lambda =3 \\ \therefore 6\lambda =24; \lambda =4 \end{gathered}$$