$解:(3)设直线CD的函数表达式是y=kx+3,把D(3,2)代入$
$得k=-\frac{1}{3}$
$即直线CD的函数表达式是y=-\frac{1}{3}x+3$
$联立\begin{cases}{y=-\frac{1}{3}+3}\\{y=x}\end{cases}，解得\begin{cases}{x=\frac{9}{4}}\\{y=\frac{9}{4}}\end{cases}$
$即点Q的坐标是(\frac{9}{4},\frac{9}{4})$
