 Hello friends, let's discuss the following question. It says, find the equation of the plane which contains the line of intersection of the planes vector r dot i cap plus 2 j cap plus 3 k cap minus 4 is equal to 0 and vector r dot 2 i cap plus j cap minus k cap plus 5 is equal to 0 and which is perpendicular to the plane vector r dot 5 i cap plus 3 j cap minus 6 k cap plus 8 is equal to 0. Now if we are given two planes vector r dot vector n1 is equal to d1 and the plane vector r dot vector n2 is equal to d2. Then equation of plane section of the two given planes is given by n1 minus d1 plus lambda into vector r dot vector n2 minus d2 is equal to 0 where lambda is any real number and we also need to know that if we have two planes vector r dot n1 equal to d1 and vector r dot n2 is equal to d2 then the perpendicular is equal to 0. So this knowledge will work as key idea behind this question. Now move on to the solution. The two given planes are vector r dot i cap plus 2 j cap minus 4 is equal to 0 and vector j cap minus k is 5 is equal to 0. Let's call this as 1 and this as 2. Now we have to find the equation of plane through the intersection of these two planes. So equation of plane section 2 j cap k cap minus 4 plus lambda into vector r dot 5 lambda is the perpendicular which is vector n2 is equal to 0 and this again implies lambda is lambda j cap minus 4 plus 5 lambda. Simply find this we get vector r is equal to vector r dot 33 i cap 5 j cap minus 41 is equal to 0. This is the position vector of any arbitrary point on the plane. So let y j cap 45 j cap 41 is equal to 0. Now taking the dot product of these two vectors we have 33 x 41 is equation. Bye for now. Take care. Have a good day.