1. Use the anti-commutative property of cross product:

$\vec P \times \vec Q = - (\vec Q \times \vec P)$

But given in the question:

$\vec P \times \vec Q = \vec Q \times \vec P$

2. Comparing both relations:

$\vec P \times \vec Q = - (\vec P \times \vec Q)$

So,

$2(\vec P \times \vec Q)=0$

Hence,

$\vec P \times \vec Q = 0$

3. Magnitude of cross product is:

$|\vec P \times \vec Q| = |\vec P|\,|\vec Q|\sin\theta$

Therefore,

$|\vec P|\,|\vec Q|\sin\theta = 0$

Assuming $\vec P$ and $\vec Q$ are non-zero vectors, we get:

$\sin\theta = 0$

4. For $0^\circ < \theta < 360^\circ$,

$\sin\theta=0 \implies \theta = 180^\circ$

($\theta=0^\circ, 360^\circ$ are not allowed in the given interval.)

Therefore, the angle is:

$\boxed{180^\circ}$

5. Comparison with stored answer:

Stored correct answer = $180$

This matches the derived answer.