 Hi and welcome to the session. Let us discuss the following question. The question says find the Cartesian as well as vector equation of the planes passing through the intersection of planes vector r dot 2i cap plus 60 cap plus 12 equals to 0 and vector r dot 3i cap plus j cap plus 4k cap equal to 0 which are at a unit distance from the origin. In this question we should know that if equation of planes r vector t1 dot vector n1 equals to d1 and vector t dot vector n2 equals to d2 let us name this as 1 and this as 2 then equation of plane passing through intersection of 1 and 2 the t n1 plus lambda times vector n2 equals to d1 plus lambda times d2. So keeping this in mind let us now begin with this illusion. Now here in this equation of plane we have 3i cap minus j cap plus 4k cap. Now given equation of planes r vector r dot 2i cap plus 60 cap plus 12 equals to 0 and vector r dot 3i cap minus j cap plus 4k cap equals to 0. Comparing this equation with this equation we find that vector n1 is equal to 2i cap plus 60 cap and vector n2 is equal to 3i cap minus j cap plus 4k cap. Now we can write this vector as 2i cap plus 6k cap plus 0k cap right. d1 is equal to minus 12 and d2 is equal to 0. Now vector equation of plane passing through intersection of given planes n1 plus lambda times vector n2 equals to d1 plus lambda times d2. This implies vector r dot vector n1 is 2i cap plus 6k cap plus 0k cap and vector n2 is 3i cap minus j cap plus 4k cap. d1 is minus 12 and d2 is 0. Now this implies vector r dot 2 plus 3 lambda i cap plus 6 minus lambda j cap plus 0 plus 4 lambda k cap is equal to minus 12. If vector equation is of the form vector r dot vector n equals to d then distance of this plane from the origin is given by mod d pi magnitude of vector n. Now here this equation of plane is also of the form vector r dot vector n equals to d. So by using this distance of this plane from the origin is given by mod of minus 12 pi square root of magnitude of vector n that is plus 3 lambda whole square plus 6 minus lambda whole square plus whole lambda whole square. Now we are given that this distance is equal to 1 plus l is equal to square root of 2 plus 3 lambda whole square plus 6 minus lambda whole square plus whole lambda whole square. Now on squaring both sides of this equation we get 144 equals to 2 plus 3 lambda whole square plus 6 minus lambda whole square plus 4 lambda whole square. Now this implies 144 is equal to 4 plus 9 lambda square plus 12 lambda plus 36 plus lambda square minus 12 lambda plus 16 lambda square. Now this implies 26 lambda square plus 40 is equal to 144 and this implies 26 lambda square plus 40 minus 144 is equal to 0. This implies 26 lambda square minus 104 is equal to 0. This implies 26 lambda square is equal to 104 and this implies lambda square is equal to 4 and this implies lambda is equal to plus minus 2. Now let's name this equation as equation number 1. Now where lambda is equal to plus 2 equation 1 becomes vector r dot 2 plus 3 into 2 i cap plus 6 minus 2 j cap plus 4 into 2 k cap equals to minus 12. Now this implies vector r dot 8 i cap plus 4 j cap plus 8 k cap is equal to minus 12. Now it is equal to minus 2 then 1 becomes vector r plus 3 into minus 2 i cap plus 6 minus minus 2 j cap plus 4 into minus 2 k cap equals to minus 12. This implies vector r dot minus 4 i cap plus 8 j cap minus 8 k cap is equal to minus 12. So these are the two vector equations of the plane which are passing through intersection of two given planes. We will find partition equation of planes is equal to x i cap plus y j cap plus z k cap. Let's name this as equation A and this as equation B. Now by taking vector r as x i cap plus y j cap plus z k cap equation A becomes x i cap plus y j cap plus z k cap dot 8 i cap 4 j cap equals to minus 12. This implies 8 x plus 4 i plus 8 z is equal to minus 12 and this implies 2 x plus 5 plus 2 z is equal to minus 3. So this is the first partition equation of the plane. Now we will find the second equation. Now by taking vector r as x i cap plus y j cap plus z k cap in equation B we get A plus y j cap plus z k cap dot minus 4 i cap plus 8 j cap equals to minus 12. Now this implies minus 4 x plus 8 y minus 8 z is equal to minus 12 and this implies minus x plus 2 y minus 2 z is equal to minus 3 which is passing through intersection of two given planes is vector r plus 4 j cap plus 8 k cap equals to minus 12 and vector 8 j cap minus 8 a cap equals to minus 12 and partition equations 2 z equals to minus 3 and minus x plus 2 y minus 2 z equals to minus 3 or we can say that 2 x plus 5 plus 2 z plus 3 equals to 0 and minus x plus 2 minus 2 z plus 3 equals to 0. So this can be the situation i and take care.