(a) Since Q is the midpoint of PR, we have:

\(\mathbf{q} = \frac{\mathbf{p} + \mathbf{r}}{2}\)

Solving for \(\mathbf{r}\):

\(2\mathbf{q} = \mathbf{p} + \mathbf{r}\)

\(\mathbf{r} = 2\mathbf{q} - \mathbf{p}\)

(b) The magnitude of the vector \(6\mathbf{i} + a\mathbf{j} + b\mathbf{k}\) is 21, so:

\(6^2 + a^2 + b^2 = 21^2\)

\(36 + a^2 + b^2 = 441\)

\(a^2 + b^2 = 405\)

The vector is perpendicular to \(3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\), so:

\(6 \times 3 + a \times 2 + b \times 2 = 0\)

\(18 + 2a + 2b = 0\)

\(2a + 2b = -18\)

\(a + b = -9\)

Substitute \(b = -9 - a\) into \(a^2 + b^2 = 405\):

\(a^2 + (-9-a)^2 = 405\)

\(a^2 + (81 + 18a + a^2) = 405\)

\(2a^2 + 18a + 81 = 405\)

\(2a^2 + 18a - 324 = 0\)

\(a^2 + 9a - 162 = 0\)

Solving the quadratic equation:

\(a = \frac{-9 \pm \sqrt{9^2 + 4 \times 162}}{2}\)

\(a = 9 \text{ or } a = -18\)

Thus, \(b = -18 \text{ or } b = 9\).