If $x < 3$, then
$\frac{x-3}{|x-3|}+\mathrm{a}=\frac{x-3}{-(x-3)}+\mathrm{a}=\mathrm{a}-1$
If $x>3$, then $\frac{|x-3|}{x-3}+\mathrm{b}=\frac{x-3}{x-3}+\mathrm{b}=1+\mathrm{b}$
$\therefore \quad$ Given function can be written as
$\mathrm{f}(x)=\left\{\begin{array}{l}
\mathrm{a}-1, x < 3 \\
\mathrm{a}+\mathrm{b}, x=3 \\
1+\mathrm{b}, x>3
\end{array}\right.$
As $\mathrm{f}(x)$ is continuous at $x=3$, we get
$\begin{array}{ll}
& \lim _{x \rightarrow 3^{-}} \mathrm{f}(x)=\mathrm{f}(3) \text { and } \lim _{x \rightarrow 3^{+}} \mathrm{f}(x)=\mathrm{f}(3) \\
\therefore \quad & \mathrm{a}-1=\mathrm{a}+\mathrm{b} \quad \text { and } 1+\mathrm{b}=\mathrm{a}+\mathrm{b} \\
\therefore \quad & \mathrm{b}=-1 \quad \text { and } \mathrm{a}=1 \\
\therefore \quad & \mathrm{a}-\mathrm{b}=2
\end{array}$