If $$(\overline{a} \times \overline{b}) \times \overline{c} = \overline{a} \times (\overline{b} \times \overline{c})$$, where $$\overline{a}, \overline{b}$$ and $$\overline{c}$$ are any three vectors such that $$\overline{a} \cdot \overline{b} \ne 0, \overline{b} \cdot \overline{c} \ne 0$$, then $$\overline{a}$$ and $$\overline{c}$$ are

Recall the vector triple‐product identity
$$\overline{p}\times(\overline{q}\times\overline{r})=\; \overline{q}\,(\overline{p}\!\cdot\!\overline{r}) \;-\; \overline{r}\,(\overline{p}\!\cdot\!\overline{q})\qquad -(1)$$

Interchanging the first two vectors gives another identity:
$$(\overline{p}\times\overline{q})\times\overline{r}= \; \overline{q}\,(\overline{p}\!\cdot\!\overline{r}) \;-\; \overline{p}\,(\overline{q}\!\cdot\!\overline{r})\qquad -(2)$$

In the given equation take $$\overline{p}=\overline{a},\; \overline{q}=\overline{b},\; \overline{r}=\overline{c}$$.

Using $$-(2)$$ on the left side:
$$(\overline{a}\times\overline{b})\times\overline{c}= \;\overline{b}\,(\overline{a}\!\cdot\!\overline{c}) \;-\; \overline{a}\,(\overline{b}\!\cdot\!\overline{c})\qquad -(3)$$

Using $$-(1)$$ on the right side:
$$\overline{a}\times(\overline{b}\times\overline{c})= \;\overline{b}\,(\overline{a}\!\cdot\!\overline{c}) \;-\; \overline{c}\,(\overline{a}\!\cdot\!\overline{b})\qquad -(4)$$

The problem states $$(\overline{a}\times\overline{b})\times\overline{c}= \overline{a}\times(\overline{b}\times\overline{c})$$. Equate $$-(3)$$ and $$-(4)$$:

$$\overline{b}\,(\overline{a}\!\cdot\!\overline{c}) \;-\; \overline{a}\,(\overline{b}\!\cdot\!\overline{c}) \;=\; \overline{b}\,(\overline{a}\!\cdot\!\overline{c}) \;-\; \overline{c}\,(\overline{a}\!\cdot\!\overline{b})$$

The first terms on each side are identical and cancel, giving
$$-\;\overline{a}\,(\overline{b}\!\cdot\!\overline{c}) \;=\; -\;\overline{c}\,(\overline{a}\!\cdot\!\overline{b})$$

Multiply by $$-1$$:
$$\overline{a}\,(\overline{b}\!\cdot\!\overline{c}) \;=\; \overline{c}\,(\overline{a}\!\cdot\!\overline{b})\qquad -(5)$$

Because $$\overline{a}\!\cdot\!\overline{b}\neq0$$ and $$\overline{b}\!\cdot\!\overline{c}\neq0$$ (given), divide both sides of $$-(5)$$ by the non-zero scalar $$\overline{a}\!\cdot\!\overline{b}$$:

$$\overline{c}= \overline{a}\;\frac{\overline{b}\!\cdot\!\overline{c}}{\overline{a}\!\cdot\!\overline{b}}$$

The right side is a scalar multiple of $$\overline{a}$$, so $$\overline{c}$$ is parallel (or antiparallel) to $$\overline{a}$$. Hence $$\overline{a}$$ and $$\overline{c}$$ are parallel vectors.

Therefore the correct choice is:
Option D — parallel