The solution to this problem is as follows:

Since the bearing of Q from P is N670W and the bearing of R from P is N230E, Q is to the west of P and R is to the east of P.

The bearing of R from Q is N680E, meaning R is also to the east of Q.

This means that P, Q and R are aligned in a straight line, with P in the middle, Q to the west and R to the east.

In a straight line, the distance between the two outer points is equal to the sum of the distances of the two inner points.

In this case, PQ is given as 150m. Since P, Q and R are aligned in a straight line, PR will also be 150m.

So the answer is: PR = 150m