 Hello and welcome to the session. Let us discuss the following question. It says, find equation of plane passing through the line of intersection of the planes vector r dot i cap plus j cap plus k cap is equal to 1 and vector r dot 2 i cap plus 3 j cap minus k cap plus 4 is equal to 0 and parallel to x axis. Now if we are given two planes vector r dot vector n 1 is equal to d 1 and the plane vector r dot vector n 2 is equal to d 2, then equation of plane through intersection of 1 and 2 is given by vector r dot vector n 1 minus d 1 plus lambda into vector r dot vector n 2 minus d 2 is equal to 0. So this knowledge will work as key idea. Let us now move on to the solution. We have to find equation of plane through the intersection of these two planes and parallel to x axis. Now the equation of plane through the intersection of vector r dot i cap plus j cap plus k cap minus 1 is equal to 0 and the plane vector r dot 2 i cap plus 3 j cap minus k cap plus 4 is equal to 0 is given by vector r dot 2 i cap it is vector r dot i cap plus j cap plus k cap minus 1 plus lambda into vector r dot 2 i cap plus 3 j cap minus k cap plus 4 is equal to 0. Now this is equal to vector r dot 1 plus 2 lambda i cap plus 1 plus 3 lambda j cap plus 1 minus lambda k cap minus 1 plus 4 lambda is equal to 0. Now r is the position vector of any arbitrary point on the plane. So let us take vector r as x i cap plus y j cap plus z k cap. So this implies x i cap plus y j cap plus z k cap dot 1 plus 2 lambda i cap plus 1 plus 3 lambda j cap plus 1 minus lambda k cap minus 1 plus 4 lambda is equal to 0. Now we take the dot product of these two vectors. So we have x into 1 plus 2 lambda plus y into 1 plus 3 lambda plus z into 1 minus lambda minus 1 plus 4 lambda is equal to 0. Let us call this as 1. Now we are given that the plane is parallel to x axis. So this implies the point 1 0 0 satisfies the equation of the plane 0 into 1 plus 3 lambda plus 0 into 1 minus lambda is equal to 0. Solving this we get lambda to be equal to minus 1 upon 2. Now put lambda is equal to minus 1 upon 2 in 1. So we have x into 1 plus 2 into minus 1 upon 2 plus y into 1 plus 3 into minus 1 upon 2 plus z into 1 minus minus 1 upon 2 minus 1 plus 4 into minus 1 upon 2 is equal to 0. So now this is equal to x into 1 minus 1 plus y into 2 minus 3 upon 2 plus z into 2 plus 1 upon 2 minus 1 minus 2 is equal to 0. Now this is equal to minus 1 by 2 y plus 3 by 2 z minus 3 is equal to 0. Taking LCM we have minus y plus 3 z minus 6 is equal to 0 and taking minus common we have y minus 3 z plus 6 is equal to 0 and this is the required equation. Hence the required equation of the plane is y minus 3 z plus 6 is equal to 0 and this completes the question. Bye for now. Take care. Have a good day.