Using the distance formula between two points in space, we get |AB|=7, |BC|=7√2, |AC|=7, AB? +AC？ =BC？ Triangle ABC is a right-angled triangle, |AB|=|AC|=7, so it is an isosceles right-angled triangle.

In case of vector calculation, then:

AB={6，-2，-3}，CB={8，-5，3}，AC={-2，3，-6}

∴|AB|=7,|AC|=7,|AB|=|AC|=7,AB.ac=6*(-2)+(-2)*3+(-3)*(-6)=0,ab⊥ac，

∴∠a = 90°, and the triangle ABC is an isosceles right triangle.