1. Use the cross-product condition

Given $\vec b \times \vec c = \vec b \times \vec d$ we get $\vec b \times (\vec c-\vec d)=\vec 0.$

This implies that $\vec c-\vec d$ is parallel to $\vec b$. Hence there exists a scalar $\lambda$ such that $\vec c-\vec d=\lambda \vec b$ or equivalently, $\vec d=\vec c-\lambda \vec b.$

2. Use the dot-product condition

We are also given $\vec a\cdot \vec d=0.$ Substitute $\vec d=\vec c-\lambda \vec b$: $\vec a\cdot(\vec c-\lambda \vec b)=0.$ So, $\vec a\cdot\vec c-\lambda(\vec a\cdot\vec b)=0.$ Therefore, $\lambda=\frac{\vec a\cdot\vec c}{\vec a\cdot\vec b}.$

Since $\vec a$ and $\vec b$ are not perpendicular, we have $\vec a\cdot\vec b\neq 0,$ so this division is valid.

3. Substitute back

Thus, $\vec d=\vec c-\left(\frac{\vec a\cdot\vec c}{\vec a\cdot\vec b}\right)\vec b.$

4. Match with the options

This is exactly Option C: $\boxed{\vec d=\vec c-\left(\frac{\vec a\cdot\vec c}{\vec a\cdot\vec b}\right)\vec b}.$

5. Check options briefly

• A has a plus sign, which does not satisfy the dot-product condition in general.
• B and D are of a different form and do not follow from $\vec b\times(\vec c-\vec d)=0$.

Hence the correct answer is C.