 Hi and welcome to the session. I am Shashin and I am going to help you with the following question. Question says, find the area of the triangle with vertices A, 1, 1, 2, B, 2, 3, 5 and C, 1, 5, 5. First of all let us understand that the area of a triangle with adjacent sides vector A and vector B is given by multiplied by magnitude of A vector cross B vector. This is the key idea to solve the given question. Let us now start with the solution. Now we are given a triangle A, B, C and coordinates of point A are 1, 1, 2, coordinates of point B are 2, 3, 5 and coordinates of point C are 1, 5, 5. Now if i, j and k are unit vectors along x, y and z axis then position vector of point A is i plus j plus 2k. Now position vector of B is given by i plus 3j plus 5k. Similarly position vector of point C is given by i plus 5j plus 5k. Now first of all we will find out vector A, B. Vector A, B represents one of the sides of the given triangle. Now we know vector A, B is equal to position vector of B minus position vector of A. Now we know position vector of B is equal to 2i plus 3j plus 5k and here we will write this minus sign as it is. Now position vector of A is i plus j plus 2k. Now subtracting corresponding components of these two vectors we get i plus 2j plus 3k is equal to vector A, B. Now we will find out adjacent side of vector A, B. Adjacent side of vector A, B is vector B, C and it is equal to position vector of C minus position vector of V. Now we know position vector of C is i plus 5j plus 5k. We will write this minus sign as it is and here we will write position vector of B. Position vector of B is 2i plus 3j plus 5k. Now subtracting the corresponding components of these two vectors we get minus i plus 2j is equal to vector B, C. Now we know area of triangle ABC is equal to half multiplied by magnitude of AB vector cross BC vector. We have already shown in key idea that area of a triangle is equal to half multiplied by magnitude of cross product of adjacent side of a triangle. Now first of all we will find out vector AB cross vector BC it is equal to determinant of unit vector i unit vector j unit vector k 1 2 3 minus 1 2 0. Now expanding this determinant with respect to this row we get unit vector i multiplied by 0 minus 6 minus unit vector j multiplied by 0 plus 3 plus unit vector k multiplied by 2 plus 2. Now this is further equal to minus 6i minus 3j plus 4k. So we get AB vector cross BC vector is equal to minus 6i minus 3j plus 4k. Now we will find out magnitude of AB vector cross BC vector it is equal to square root of minus 6 square plus minus 3 square plus 4 square. Now this is further equal to square root of 36 plus 9 plus 16. Now adding these three terms we get square root of 61. So magnitude of AB vector cross BC vector is equal to square root of 61. Now we get area of triangle ABC is equal to half multiplied by square root of 61. We know magnitude of AB vector cross BC vector is equal to square root of 61. So here we can write square root of 61 and we get area of triangle ABC is equal to half multiplied by square root of 61. This completes the session hope you understood the solution take care and have a nice day.