 Hi, and welcome to the session. Let us discuss the following question. The question says find x, such that the four points a having coordinates 3 to 1, b having coordinates 4x5, c, 4, 2, minus 2, and b, 6, 5, minus 1 are coplanar. Before solving this question, we should know that the points a, b, c, and d are coplanar if the vectors vector a, b, vector a, c, and vector a, d are coplanar. We know that three vectors are coplanar if their scalar triple product is equal to 0. So vector a, b, vector a, c, and a, d are coplanar if their scalar triple product is equal to 0. This is our key idea in this question. Let's now begin with the solution. The given points are a having coordinates 3 to 1, b having coordinates 4x5, c having coordinates 4, 2, minus 2, and d having coordinates 6, 5, minus 1. Let the origin of these points be o. We can write the position vectors of the four points with respect to the origin as vector oa is equal to 3i cap plus 2j cap plus k cap. Vector ob as 4i cap plus xj cap plus 5k cap. Vector oc as 4i cap plus 2j cap minus 2k cap. And vector od as 6i cap plus 5j cap minus k cap. Now we have to find vector a, b, vector a, c, and vector a, d. Now vector ab is equal to vector ob minus vector oa. Vector ob is equal to 4i cap plus xj cap plus 5k cap. And vector oa is equal to 3i cap plus 2j cap plus k cap. And this is equal to 4 minus 3 is 1. So we have i cap plus x minus 2j cap plus 5 minus 1 So we have 4k cap plus vector ab is equal to i cap plus x minus 2j cap plus 4k cap. Now we will find vector ac. Vector ac is equal to vector oc minus vector oa. And this is equal to i cap minus 3k cap. Now lastly we will find vector ad. Vector ad is equal to vector od minus vector oa. And this is equal to 3i cap plus 3j cap minus 2k cap. Now according to our key idea, the points ab, c, and d are co-planar. If their scalar triple product is 0, that is scalar triple product of vector ab, vector ac, and vector ad is equal to 0. Now this implies determinant of 1x minus 2 for 1, 0 minus 3, 3, 3 minus 2 is equal to 0. And this implies 1 into determinant of 0, 3 minus 3 minus 2, minus x minus 2 into determinant of 1, 3 minus 3 minus 2, plus 4 into determinant of 1, 3, 0, 3 is equal to 0. And this implies 1 into 9 minus x minus 2 into 7, plus 4 into 3 is equal to 0. And this implies 9 minus 7x plus 14 plus 12 is equal to 0. And this implies minus 7x plus 35 is equal to 0. And this implies x is equal to 5. Hence the required value of x is 5. This completes the session. Bye and take care.