$\mathrm{A}(3, 4), \mathrm{B}(2, 5), \mathrm{C}(3, 7) then ∆ABC is$ 

Given points 

$$\begin{gathered} A(3, 4), \mathrm{B}=(2, 5), \mathrm{C}=(3, 7) \\ \mathrm{AB}=\sqrt{{(-1)}^2+{\left(1\right)}^2}=\sqrt{2}\Rightarrow {\mathrm{AB}}^2=2 \\ \mathrm{BC}=\sqrt{{\left(1\right)}^2+{\left(2\right)}^2}=\sqrt{5}\Rightarrow {\mathrm{BC}}^2=5 \\ {\mathrm{CA}}^2=\sqrt{{\left(0\right)}^2+{\left(3\right)}^2}=\sqrt{9}\Rightarrow {\mathrm{CA}}^2=9 \\ \mathrm{Here} {\mathrm{AB}}^2+{\mathrm{BC}}^2 < {\mathrm{CA}}^2 \end{gathered}$$

$\therefore$ Triangle ABC is an obtuse angled  scalane triangle.