jmc

Since $2x+7$ is a factor, we should get a remainder of $0$ when we divide $6x^3+19x^2+cx+35$. $$\begin{array}{ccccc}\text{multicolumn}2r3x^2& -x& +5& & \\ \text{cline}2-52x+7& 6x^3& +19x^2& +cx& +35\\ \text{multicolumn}2r-6x^3& -21x^2& & & \\ \text{cline}2-3\text{multicolumn}2r0& -2x^2& +cx& & \\ \text{multicolumn}2r& +2x^2& +7x& & \\ \text{cline}3-4\text{multicolumn}2r& 0& (c+7)x& +35& \\ \text{multicolumn}2r& & -10x& -35& \\ \text{cline}4-5\text{multicolumn}2r& & (c+7-10)x& 0& \end{array}$$The remainder is $0$ if $c+7-10=0$, so $c=\boxed{3}$.