Everything is defined at the given point, so we can just directly substitute.

\(\displaystyle\lim_{{{\left({x},{y},{z}\right)}\rightarrow{\left(-{3},{1},{2}\right)}}}{\frac{{{\ln{{z}}}}}{{{x}{y}-{z}}}}={\frac{{{\ln{{2}}}}}{{-{3}{\left({1}\right)}-{2}}}}=-{\frac{{{\ln{{2}}}}}{{{5}}}}\)

The function is continuous at this point and where \(\displaystyle{x}{y}\ne{z}\)

\(\displaystyle\lim_{{{\left({x},{y},{z}\right)}\rightarrow{\left(-{3},{1},{2}\right)}}}{\frac{{{\ln{{z}}}}}{{{x}{y}-{z}}}}={\frac{{{\ln{{2}}}}}{{-{3}{\left({1}\right)}-{2}}}}=-{\frac{{{\ln{{2}}}}}{{{5}}}}\)

The function is continuous at this point and where \(\displaystyle{x}{y}\ne{z}\)