jmc

Instead of plugging in $x=7$ into $f(x)$ and solving, we can use the Remainder Theorem to avoid complicated arithmetic. We know that $f(7)$ will be the remainder when $f(x)$ is divided by $x-7$. So we have:

$$\begin{array}{cccccc}\text{multicolumn}2r2x^3& -3x^2& +5x& +11& & \\ \text{cline}2-6x-7& 2x^4& -17x^3& +26x^2& -24x& -60\\ \text{multicolumn}2r2x^4& -14x^3& & & & \\ \text{cline}2-3\text{multicolumn}2r0& -3x^3& +26x^2& & & \\ \text{multicolumn}2r& -3x^3& +21x^2& & & \\ \text{cline}3-4\text{multicolumn}2r& 0& 5x^2& -24x& & \\ \text{multicolumn}2r& & 5x^2& -35x& & \\ \text{cline}4-5\text{multicolumn}2r& & 0& 11x& -60& \\ \text{multicolumn}2r& & & 11x& -77& \\ \text{cline}5-6\text{multicolumn}2r& & & 0& 17& \end{array}$$Hence $f(7)=\boxed{17}$.