Given, triangle whose vertices are at A = (3, -1, 2)
=3−+2 B=(1,−1,−3)=−−3 and
C=(4,−3,1)=4−3+ Let A, B , and C be the vertices of the △ABC , then the area of that triangle
=×|×| =(−−3)−(3−+2) ⇒=−2+0−5 =(4−3+)−(3−+2) ⇒=−2− Now,
×=|| ⇒×=(−10)−(2+5)+(4−0) ⇒×=−10−7+4 ⇒|×|=√(−10)2+(−7)2+(4)2 ⇒|×|=√165 The area of that triangle
=×|×| ⇒ The area of that triangle
=×√165 Hence, the area of a triangle whose vertices are at (3, -1, 2), (1, -1, -3) and (4, -3, 1) is