jmc

We use the distance formula: $$\sqrt{(7-2{)}^2+(15-3{)}^2}=\sqrt{25+144}=\boxed{13}.$$

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We note that the points $(2,3)$, $(7,15)$, and $(7,3)$ form a right triangle with legs of length 5 and 12. This is a Pythagorean triple, so the hypotenuse has length $\boxed{13}$.