We need the scalar projection of $\overrightarrow{BA}$ on $\overrightarrow{BC}$.

That is $\text{proj}_{\overrightarrow{BC}}(\overrightarrow{BA})=\frac{\overrightarrow{BA}\cdot\overrightarrow{BC}}{|\overrightarrow{BC}|}.$

Given side lengths: $|\overrightarrow{BC}|=3,\quad |\overrightarrow{CA}|=5,\quad |\overrightarrow{BA}|=7.$

In triangle $ABC$, the vectors satisfy $\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{BA}.$

Taking modulus squared: $|\overrightarrow{BA}|^2=|\overrightarrow{BC}+\overrightarrow{CA}|^2.$ So, $49=9+25+2\,\overrightarrow{BC}\cdot\overrightarrow{CA}.$ Hence, $49=34+2\,\overrightarrow{BC}\cdot\overrightarrow{CA}$ $2\,\overrightarrow{BC}\cdot\overrightarrow{CA}=15$ $\overrightarrow{BC}\cdot\overrightarrow{CA}=\frac{15}{2}.$

Now, $\overrightarrow{BA}=\overrightarrow{BC}+\overrightarrow{CA}.$ Dot both sides with $\overrightarrow{BC}$: $\overrightarrow{BA}\cdot\overrightarrow{BC}=\overrightarrow{BC}\cdot\overrightarrow{BC}+\overrightarrow{CA}\cdot\overrightarrow{BC}.$ Therefore, $\overrightarrow{BA}\cdot\overrightarrow{BC}=|\overrightarrow{BC}|^2+\overrightarrow{CA}\cdot\overrightarrow{BC}$ $=9+\frac{15}{2}=\frac{33}{2}.$

Hence the projection is

Therefore, the correct option is $\boxed{\text{C }\frac{11}{2}}.$