Suppose  a, b, c are such that the curve $\mathrm{y}={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}$ is $\mathrm{y}=3\mathrm{x}-3 \mathrm{at} \left(1,0\right)$ and is also tangent to $\mathrm{y}=\mathrm{x}+1 \mathrm{at} \left(3,4\right)$ Then the value of $\left(2\mathrm{a}-\mathrm{b}-4\mathrm{c}\right)$ equals _____

$\begin{array}{l}\mathrm{y}={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}, \frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{ax}+\mathrm{b}\\ \mathrm{When} \mathrm{x}=1, \mathrm{y}=0\\ \therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=0 \left(1\right)\\ {\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|}_{\mathrm{x}=1}=3 \mathrm{and} {\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|}_{\mathrm{x}=3}=1\\ 2\mathrm{a}+\mathrm{b}=3 \left(2\right)\\ 6\mathrm{a}+\mathrm{b}=1 \left(3\right)\\ \mathrm{Solving} \left(1\right), \left(2\right) \mathrm{and} \left(3\right),\\ \mathrm{a}=-\frac{1}{2}, \mathrm{b}=4, \mathrm{c}=-\frac{7}{2}\\ \therefore 2\mathrm{a}-\mathrm{b}-4\mathrm{c}=-1-4+14=9\end{array}$