To solve this problem, we resolve the forces into components along and perpendicular to the direction of PQ.

(i) (a) For resolving in the direction PQ:

The component is given by:

\(2 \times 10 \cos 30^\circ - 6 \cos 60^\circ\)

\(= 20 \times \frac{\sqrt{3}}{2} - 6 \times \frac{1}{2}\)

\(= 10\sqrt{3} - 3\)

\(\approx 14.3 \text{ N}\)

(i) (b) For resolving perpendicular to PQ:

The component is given by:

\(\pm 6 \cos 30^\circ - 6 \cos 60^\circ\)

\(= \pm 6 \times \frac{\sqrt{3}}{2} - 6 \times \frac{1}{2}\)

\(= \pm 3\sqrt{3} - 3\)

\(\approx \pm 5.20 \text{ N}\)

(ii) To find the magnitude of the resultant of the three forces:

Use the formula:

\(\sqrt{(\text{component in direction of } PQ)^2 + (\text{component perpendicular to } PQ)^2}\)

\(= \sqrt{(14.3)^2 + (5.20)^2}\)

\(\approx 15.2 \text{ N}\)