First, convert the vector the plane is orthogonal to to standard notation.

\(\displaystyle{2}{i}+{3}{j}-{k}={<}{2},{3},-{1}{>}\)

Plug the vector from the step above and the point we were given into the standard equation of a plane in space

\(\displaystyle{2}{\left({x}-{3}\right)}+{3}{\left({y}-{2}\right)}-{1}{\left({z}-{2}\right)}={0}\)

Simplify the equation from the step above. This is the equation of the plane asked for in the question.

\(\displaystyle{2}{x}+{3}{y}-{z}={10}\)

Result: \(\displaystyle{2}{x}+{3}{y}-{z}={10}\)