Let P(0,7,10),Q(−1,6,6) and R(−4,9,6) be the vertices of a triangle Here, PQ=√1+1+16=3√2 QR=√9+9+0=3√2 PR=√16+4+16=6 Now, PQ2+QR2=(3√2)2+(3√2)2=36=(PR)2 Therefore,Δ‌PQR is a right angled triangle at Q. Also, OQ=QR. Hence, Δ‌PQR is a right angled isosceles triangle.