(a) Write

\[ \overrightarrow{OA}=\begin{pmatrix}2\\2\\-1\end{pmatrix}, \quad \overrightarrow{OB}=\begin{pmatrix}4\\2\\4\end{pmatrix}. \]

The perpendicular distance from \(A\) to the line through \(O\) and \(B\) is

\[ \frac{|\overrightarrow{OA}\times\overrightarrow{OB}|}{|\overrightarrow{OB}|}. \]

Now

\[ \overrightarrow{OA}\times\overrightarrow{OB} = \begin{pmatrix}2\\2\\-1\end{pmatrix} \times \begin{pmatrix}4\\2\\4\end{pmatrix}. \]

\[ = \begin{pmatrix} 10\\ -12\\ -4 \end{pmatrix}. \]

So

\[ |\overrightarrow{OA}\times\overrightarrow{OB}| = \sqrt{10^2+(-12)^2+(-4)^2} =\sqrt{260}=2\sqrt{65}. \]

Also,

\[ |\overrightarrow{OB}|=\sqrt{4^2+2^2+4^2}=\sqrt{36}=6. \]

Hence the perpendicular distance is

\[ \frac{2\sqrt{65}}{6}=\frac13\sqrt{65}. \]

(b) We have

\[ \overrightarrow{OC}=\begin{pmatrix}3\\p\\q\end{pmatrix}. \]

Also

\[ \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA} = \begin{pmatrix}4\\2\\4\end{pmatrix} - \begin{pmatrix}2\\2\\-1\end{pmatrix} = \begin{pmatrix}2\\0\\5\end{pmatrix}. \]

Since \(\overrightarrow{OC}\) is perpendicular to \(\overrightarrow{AB}\),

\[ \overrightarrow{OC}\cdot\overrightarrow{AB}=0. \]

So

\[ \begin{pmatrix}3\\p\\q\end{pmatrix} \cdot \begin{pmatrix}2\\0\\5\end{pmatrix} =0. \]

\[ 6+5q=0. \]

Therefore

\[ q=-\frac65. \]

Since angle \(AOC=\) angle \(COB\),

\[ \frac{\overrightarrow{OA}\cdot\overrightarrow{OC}}{|\overrightarrow{OA}|\,|\overrightarrow{OC}|} = \frac{\overrightarrow{OC}\cdot\overrightarrow{OB}}{|\overrightarrow{OC}|\,|\overrightarrow{OB}|}. \]

Cancel \(|\overrightarrow{OC}|\):

\[ \frac{\overrightarrow{OA}\cdot\overrightarrow{OC}}{|\overrightarrow{OA}|} = \frac{\overrightarrow{OC}\cdot\overrightarrow{OB}}{|\overrightarrow{OB}|}. \]

Now

\[ |\overrightarrow{OA}|=\sqrt{2^2+2^2+(-1)^2}=3 \]

and

\[ |\overrightarrow{OB}|=6. \]

So

\[ 2(\overrightarrow{OA}\cdot\overrightarrow{OC}) = \overrightarrow{OC}\cdot\overrightarrow{OB}. \]

Now

\[ \overrightarrow{OA}\cdot\overrightarrow{OC} = 2(3)+2p+(-1)q=6+2p-q. \]

Using \(q=-\frac65\),

\[ \overrightarrow{OA}\cdot\overrightarrow{OC}=6+2p+\frac65=\frac{36}{5}+2p. \]

Also

\[ \overrightarrow{OC}\cdot\overrightarrow{OB} = 3(4)+2p+4q=12+2p+4q. \]

Using \(q=-\frac65\),

\[ \overrightarrow{OC}\cdot\overrightarrow{OB}=12+2p-\frac{24}{5}=\frac{36}{5}+2p. \]

Therefore

\[ 2\left(\frac{36}{5}+2p\right)=\frac{36}{5}+2p. \]

\[ \frac{72}{5}+4p=\frac{36}{5}+2p. \]

\[ 2p=-\frac{36}{5}. \]

So

\[ p=-\frac{18}{5}. \]

Answer: \(p=-\frac{18}{5}\), \(q=-\frac65\).