1. Let $f(x)=ax^2+bx+c.$ Given: $f(1)=3,\\ f(-2)=\lambda,\\ f(3)=4.$ Also, $f(0)+f(1)+f(-2)+f(3)=14.$

2. Substitute the known values into the sum: $f(0)+3+\lambda+4=14.$ Since $f(0)=c$, we get $c+\lambda+7=14$ $c+\lambda=7 \quad \Rightarrow \quad \lambda=7-c.$

3. Now use the conditions $f(1)=3$ and $f(3)=4$.

From $f(1)=3$: $a+b+c=3. \qquad (1)$

From $f(3)=4$: $9a+3b+c=4. \qquad (2)$

Subtract (1) from (2): $8a+2b=1$ $4a+b=\frac12. \qquad (3)$

4. Compute $f(-2)$: $\lambda=f(-2)=4a-2b+c.$ Using $c=7-\lambda$ from step 2, $\lambda=4a-2b+(7-\lambda)$ $2\lambda=4a-2b+7$ $\lambda=2a-b+\frac72. \qquad (4)$

5. From (3), $b=\frac12-4a.$ Substitute into (4): $\lambda=2a-\left(\frac12-4a\right)+\frac72$ $\lambda=2a-\frac12+4a+\frac72$ $\lambda=6a+3.$

Now use (1): $a+b+c=3.$ Substitute $b=\frac12-4a$: $a+\frac12-4a+c=3$ $-3a+c=\frac52$ $c=3a+\frac52.$

But from step 2, $c=7-\lambda=7-(6a+3)=4-6a.$ So, $3a+\frac52=4-6a$ $9a=\frac32$ $a=\frac16.$

Then $\lambda=6a+3=6\cdot\frac16+3=1+3=4.$

6. Therefore, the correct option is $\boxed{4}.$ So, option D is correct.