Answer: (-1/3, 2/3)
The circumcentre P(x, y) is equidistant from A(-3, 1), B(0, -2), C(1, 3).
\(PA^2=PB^2\): \((x+3)^2+(y-1)^2=x^2+(y+2)^2\)
\[6x+10-2y=4y+4\ \Rightarrow\ x-y=-1\quad\text{(I)}\]
\(PB^2=PC^2\): \(x^2+(y+2)^2=(x-1)^2+(y-3)^2\)
\[4y+4=-2x+1-6y+9\ \Rightarrow\ x+5y=3\quad\text{(II)}\]
Subtract (I) from (II): \(6y=4\), so \(y=\dfrac23\) and \(x=y-1=-\dfrac13\).