Quite the same

Since $\triangle PQR$ is an equilateral triangle, we have $RQ=PQ=PC+CQ=3+2=5$ . Let $AR=x$ . By cosine rule, we have:

$\begin{cases} AB^2 = BR^2+AR^2-2BR\dot{}AR \cos 60^\circ = 3^2 +x^2 - 3x & = x^2 - 3x + 9 \\ BC^2 = BP^2+PC^2-2BP\dot{}PC \cos 60^\circ = 2^2+3^2-2(3) & = 7 \\ AC^2 = AQ^2 + CQ^2 -2AQ\dot{}CQ \cos 60^\circ = (5-x)^2+2^2-2(5-x) & = x^2 - 8x + 19 \end{cases}$

By Pythagorean theorem, we have $BC^2 +AC^2 = AB^2$ :

$\begin{aligned} x^2 - 3x + 9 & = 7 + x^2 - 8x + 19 \\ \Rightarrow 5x & = 17 \\ x & = \frac{17}{5} \\ \Rightarrow AQ & = 5 - x = 5 - \frac{17}{5} = \boxed{\dfrac{8}{5}} \end{aligned}$

$BC=\sqrt{2^2+3^2-2*3*2*(1/2)}$ $\frac{PB}{sin(\angle{PCB})}=\frac{BC}{sin(\angle{BPC})}$ $\angle{BCP}=asin(0.654) \rightarrow \angle{ACQ}=90-\angle{BCP}$ $\rightarrow \angle{CAQ}=120-\angle{ACQ}$ $\frac{AQ}{sin(\angle{ACQ})}=\frac{CQ}{sin(\angle{CAQ})}$ $AQ=\boxed{\frac{8}{5}}$