Answer
$x=3$ is not a vertical asymptote but the $x$-value where the hole of the graph is located.
Work Step by Step
Given $$h(x)=\frac{6-2x}{3-x}$$
Rewrite the numerator:
$$h(x)=\frac{2(3-x)}{3-x}$$ $$h(x)=2\text{, where } x\not=3$$
Equating $3-x=0$, $$x=3$$
Thus, the error is $x=3$ is not a vertical asymptote but the $x$-value where the hole of the graph is located.
