The first case: P is on AB and Q is on BC. Suppose that the area of the triangle PCQ after t seconds of movement is 12.6.

Have:? (6-t)*(8-2t)= 12.6

Solution: t= 1.3, t=8.7 (discarded)

In the second case, P is on BC and Q is on AC. Suppose it moves for t seconds. Let the height from q to BC be h.

Yes: 1/2 * PC*h= 12.6.

Because PC = 8-(t-6) =14-t.

According to the triangle similarity: h/AB=CQ/AC.

Therefore, h=3/5 *CQ.

=3/5 *(t-8)

Therefore,1/2 * 8-(t-6) * 3/5 * (t-8) =12.6.

Solved: t=5.5 (excluding), t= 12.5.

So the answer is 12.5 seconds or 1.3 seconds.