 Hello and welcome to the session. In this session we are going to discuss the following question which says that find a unit vector perpendicular to the plane A, B, C where A with the coordinates 1, 2, 3, B with the coordinates 3, 4, 5 and C with the coordinates minus 1, 6, minus 7 are the points on the plane. Let's start the solution. Let OVB origin and we are given point A with the coordinates 1, 2, 3, point B with the coordinates 3, 4, 5 and point C with the coordinates minus 1, 6, minus 7. Then we have vector OA can be written as i cap plus 2 j cap plus 3 k cap as point A has coordinates 1, 2, 3. Vector OB can be written as 3 i cap plus 4 j cap plus 5 k cap as point B has coordinates 3, 4, 5. Vector OB can be written as minus i cap plus 6 j cap minus 7 k cap as point C has coordinates minus 1, 6, minus 7. Therefore vector AB is given by vector OB minus vector OA which is equal to 3 i cap plus 4 j cap plus 5 k cap minus OA that is i cap plus 2 j cap plus 3 k cap. Therefore vector AB is equal to 2 i cap plus 2 j cap plus 2 k cap and vector AC is given by vector OC minus vector OA which is equal to minus of i cap plus 6 j cap minus 7 k cap minus of i cap plus 2 j cap plus 3 k cap. Therefore vector AC is given by minus 2 i cap plus 4 j cap minus 10 k cap. Now we shall find the cross product of the vectors AB and AC. Now vector AB into vector AC is equal to which is equal to i cap into minus 20 minus 8 minus j cap into minus 20 plus 4 plus j cap into 8 plus 4 which is equal to minus 28 i cap plus 16 j cap plus 12 k cap. Therefore the cross product of vector AB and AC is equal to minus 28 i cap plus 16 j cap plus 12 k cap and modulus of the cross product of vector AB and vector AC is given by square root of minus 28 square plus 16 square plus 12 square which is equal to square root of 784 plus 256 plus 144 which is equal to square root of 1184 that is 4 into square root of 74. Therefore modulus of the cross product of the vector AB and AC is equal to 4 into square root of 74. Now we have cross product of vector AB into AC is equal to minus 28 i cap plus 16 j cap plus 12 k cap and the modulus of the cross product of the vector AB and AC is equal to 4 into square root of 74. Now our unit vector perpendicular to the plane of triangle ABC is perpendicular to both AB and AC. Therefore our unit vector N perpendicular to the plane of triangle ABC is given by N cap is equal to vector AB into vector AC upon modulus of vector AB into vector AC which is equal to minus 28 i cap plus 16 j cap plus 12 k cap upon 4 into square root of 74 which can be written as 1 upon square root of 74 into minus 28 by 4 i cap plus 16 by 4 j cap plus 12 by 4 k cap which is equal to 1 by square root of 74 into minus 7 i cap plus 4 j cap plus 3 k cap or we can also write minus 1 upon square root of 74 into 7 i cap minus 4 j cap minus 3 k cap hence our unit vector perpendicular to the plane of triangle ABC is given by minus 1 upon square root of 74 into 7 i cap minus 4 j cap minus 3 k cap which is the required answer. This completes our session but we enjoyed this session.