Answer: \(\overrightarrow{AB}\times\overrightarrow{AC}=(1,1,-1).\)

The area of triangle \(ABC\) is \(\displaystyle \frac{\sqrt3}{2}\).

The perpendicular distance from \(C\) to \(AB\) is \(\displaystyle \sqrt{\frac{3}{14}}\).

From the coordinates in the question,

\(\overrightarrow{AB}=(1,2,3),\qquad \overrightarrow{AC}=(1,1,2).\)

Compute the cross product:

\(\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&2&3\\1&1&2\end{vmatrix}=(1,1,-1).\)

The magnitude is

\(|\overrightarrow{AB}\times\overrightarrow{AC}|=\sqrt{1^2+1^2+(-1)^2}=\sqrt3.\)

The area of the triangle is half the area of the parallelogram formed by \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\), so

\(\text{Area}=\frac12\sqrt3.\)

If \(h\) is the perpendicular distance from \(C\) to the line \(AB\), then

\(\text{Area of triangle }ABC=\frac12|\overrightarrow{AB}|h.\)

Now

\(|\overrightarrow{AB}|=\sqrt{1^2+2^2+3^2}=\sqrt{14}.\)

Hence

\(\frac{\sqrt3}{2}=\frac12\sqrt{14}\,h,\)

so

\(h=\frac{\sqrt3}{\sqrt{14}}=\sqrt{\frac{3}{14}}.\)