The vectors $$\vec{a}$$ and $$\vec{b}$$ are not perpendicular and $$\vec{c}$$ and $$\vec{d}$$ are two vectors satisfying: $$\vec{b} \times \vec{c} = \vec{b} \times \vec{d}$$ and $$\vec{a} \cdot \vec{d} = 0$$. Then the vector $$\vec{d}$$ is equal to:

Condition $$\vec{b}\times\vec{c}=\vec{b}\times\vec{d}$$ implies $$\vec{b}\times(\vec{c}-\vec{d})=\vec{0}$$.

For a non-zero vector $$\vec{b}$$, the cross product with another vector is zero only when the two vectors are parallel. Hence $$\vec{c}-\vec{d}=k\vec{b} \quad\text{for some scalar }k.$$ Therefore, $$\vec{d}=\vec{c}-k\vec{b}$$ $$-(1)$$

The second condition is $$\vec{a}\cdot\vec{d}=0$$. Substitute from $$-(1)$$:

$$\vec{a}\cdot(\vec{c}-k\vec{b})=0$$ $$\Longrightarrow\; \vec{a}\cdot\vec{c}-k\,\vec{a}\cdot\vec{b}=0$$.

Because $$\vec{a}$$ and $$\vec{b}$$ are not perpendicular, $$\vec{a}\cdot\vec{b}\neq0$$, so

$$k=\dfrac{\vec{a}\cdot\vec{c}}{\vec{a}\cdot\vec{b}}.$$

Insert this value of $$k$$ back into $$-(1)$$:

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

Thus $$\vec{d}$$ equals the expression given in Option C.

Final answer: Option C which is: $$\vec{c}-\left(\dfrac{\vec{a}\cdot\vec{c}}{\vec{a}\cdot\vec{b}}\right)\vec{b}$$.