Let ABC be a triangle, the position vectors of whose vertices are 7j^+10k^,−i^+6j^+6k^ and −4i^+9j^+6k^ then ∆ABC is

we have AB→=−i^−j^−4k^, BC→=−3i^+3j^ and CA→=4i^−2j^+4k^Therefore, |AB→|=|BC→|=32clearly ,|AB→|2+|BC→|2=|AC→|2Hence, the triangle is right-angled isosceles triangle.