 Hello and welcome to the session. In this session we are going to discuss the following question which says that sign the area of the triangle formed by the point A with the coordinates 12, 1, 5, B with the coordinates 0, 9, 1, C with the coordinates minus 4, minus 3, 9, vector area of any triangle C, A, B, C is given by 1 by 2 into cross product of vector A and B plus cross product of vector B and C plus cross product of vector C and A. With this key idea let us proceed with the solution. We are given point A with the coordinates 12, 1, 5, point B with the coordinates 0, 9, 1, point C with the coordinates minus 4, minus 3, 9, are the three vertices of triangle A, B, C. Let A, B, C be the position vectors of vertices A, B and C respectively. Then we have position vector of A is equal to 12 i cap plus j cap plus 5 k cap, position vector of B is equal to 0 i cap plus 9 j cap plus k cap, position vector of C is equal to minus 4 i cap, minus 3 j cap plus 9 k cap. Let us find cross product of vector A and B, vector A is given by 12 i cap plus j cap plus 5 k cap and vector B is given by 0 i cap plus 9 j cap plus k cap. Cross product of vector A and B is given by i cap into 1 minus 45 minus j cap into 12 minus 0 plus k cap into 108 minus 0 which is equal to minus 44 i cap minus 12 j cap plus 108 k cap. So, cross product of vector A and B is given by minus 44 i cap minus 12 j cap plus 108 k cap. Next we shall find cross product of vector B and C, vector B is given by 0 i cap plus 9 j cap plus k cap and vector C is given by minus 4 i cap minus 3 j cap plus 9 k cap. So, cross product of vector B and C is given by i cap into 81 plus 3 minus j cap into 0 plus 34 plus k cap into 0 plus 36 which is equal to 84 i cap minus 4 j cap plus 36 k cap. Therefore, cross product of vector B and C is given by 84 i cap minus 4 j cap plus 36 k cap. Now we shall find cross product of vector C and A, vector C is given by minus 4 i cap minus 3 j cap plus 9 k cap and vector A is given by 12 i cap plus j cap plus 5 k cap. Cross product of vector C and A is given by i cap into minus 15 minus 9 minus j cap into minus 20 minus 108 plus k cap into minus 4 plus 36 which is equal to minus 24 i cap plus 128 j cap plus 32 k cap. Therefore, cross product of vector C and A is given by minus 24 i cap plus 128 j cap plus 32 k cap. Now we have got the value of the cross product of vector A and B, the cross product of vector B and C and the cross product of vector C and A and from the key here we know that vector area of triangle ABC is given by 1 by 2 into cross product of vector A and B plus cross product of vector D and C plus cross product of vector C and A. Therefore, vector area of triangle ABC is given by 1 by 2 into cross product of vector A and that is minus 34 i cap minus 12 j cap plus 108 k cap plus cross product of vector B and C that is 84 i cap minus 4 j cap plus 36 k cap plus cross product of vector C and A that is minus 24 i cap plus 128 j cap plus 32 k cap which is equal to 1 by 2 into 16 i cap plus 12 j cap plus 176 k cap which is equal to 8 i cap plus 56 j cap plus 88 k cap. So, vector area of triangle ABC is given by 8 i cap plus 56 j cap plus 88 k cap therefore required area of triangle ABC is given by modulus of 8 i cap plus 56 j cap plus 88 k cap which is equal to square root of 8 square plus 56 square plus 88 square which is equal to square root of 64 plus 3136 plus 7744 which is equal to square root of 10944 which is equal to 24 into square root of 19. Therefore, the area of triangle ABC is given by 24 square root of 19 square units which is the required answer. This completes our session hope you enjoyed this session.