Answer: \(\vec{OP}=11\mathbf i-4\mathbf j+3\mathbf k\), and \(\vec{OQ}=5\mathbf i-7\mathbf j+\mathbf k\).

(i) \(\overrightarrow{AP}\times\overrightarrow{AQ}=12\mathbf i+6\mathbf j-45\mathbf k\), and the area of triangle \(APQ\) is \(\frac{\sqrt{2205}}2\).

(ii) The volume of tetrahedron \(APQB\) is \(\frac{245}{2}\).

Points on the two lines have position vectors

\(\vec{OP}=(8+\lambda,\,2-2\lambda,\,3)\), and \(\vec{OQ}=(5,\,3+2\mu,\,-14-3\mu)\).

Thus

\(\overrightarrow{PQ}=\vec{OQ}-\vec{OP}=(-3-\lambda,\,1+2\lambda+2\mu,\,-17-3\mu)\).

The direction vector of \(l_1\) is \((1,-2,0)\), and the direction vector of \(l_2\) is \((0,2,-3)\). Since \(PQ\) is perpendicular to both lines,

\((1,-2,0)\cdot\overrightarrow{PQ}=0\),

and

\((0,2,-3)\cdot\overrightarrow{PQ}=0\).

These give

\(5\lambda+4\mu=-5\),

and

\(4\lambda+13\mu=-53\).

Solving gives \(\lambda=3\) and \(\mu=-5\).

Therefore

\(\vec{OP}=(11,-4,3)=11\mathbf i-4\mathbf j+3\mathbf k\),

and

\(\vec{OQ}=(5,-7,1)=5\mathbf i-7\mathbf j+\mathbf k\).

Now \(A=(8,2,3)\) and \(B=(5,3,-14)\). Hence

\(\overrightarrow{AP}=(3,-6,0)\), and \(\overrightarrow{AQ}=(-3,-9,-2)\).

So

\(\overrightarrow{AP}\times\overrightarrow{AQ}=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\3&-6&0\\-3&-9&-2\end{vmatrix}=12\mathbf i+6\mathbf j-45\mathbf k\).

The area of triangle \(APQ\) is

\(\frac12\left|\overrightarrow{AP}\times\overrightarrow{AQ}\right|=\frac12\sqrt{12^2+6^2+45^2}=\frac{\sqrt{2205}}2\).

For the volume, use the scalar triple product. Since

\(\overrightarrow{AB}=(-3,1,-17)\),

we have

\(\overrightarrow{AB}\cdot(\overrightarrow{AP}\times\overrightarrow{AQ})=(-3,1,-17)\cdot(12,6,-45)=735\).

The volume of tetrahedron \(APQB\) is therefore

\(\frac16\left|\overrightarrow{AB}\cdot(\overrightarrow{AP}\times\overrightarrow{AQ})\right|=\frac{735}{6}=\frac{245}{2}\).