Find the vector of the plane passing through the intersection of the planes `vecr.(2hati+2hatj-3hatk)=7, vecr.(2hati+5hatj+3hatk)=9` and the point (2,1,3)`.

Find the vector of the plane passing through the intersection of the p

1.	Find the vector of the plane passing through the intersection of the planes `vecr.(2hati+2hatj-3hatk)=7, vecr.(2hati+5hatj+3hatk)=9` and the point (2,1,3)`.
Answer» Equation of given planes are `vecr.(2hati+2hatj-3hatk)=7` and `vecr.(2hati+5hatj+3hatk) = 9` The equation of these planes can be written as ` v ecr.(2hati+2hatj-3hatk) - 7 = 0 "………"(1)` and `vecr.(2hati+5hatj+3hatk) - 9 = 0 "….."(2)` Equation of a plane through the intersection of plane (1) and (2) is, `[vecr.(2hati+2hatj-3hatk)-7]+lambda[vecr.(2hati+5hatj+3hatk)-9]=0` `rArr vecr.[(2hati+2hatj-3hatk)+lambda(2hati+5hatj+3hatk)]=9lambda+7` `rArr [vecr.(2+2lambda)hati+(2+5lambda)hatj+(3lambda-3)hatk]=9lambda+7"......."(3)` This intersecting plane passes through the point (2,1,3), whose position vector is `vecr.(2hati+hatj+3hatk)`. Put this value of `vecr` in eqaution (3), `(2hati+hatj+ 3hatk) .[(2+2lambda) hati+(2+5lambda)hatj+(3lambda-3)hatk] = 6 lambda + 7` `rArr 2(2+2lambda)+(2+5lambda)+3(3lambda-3)=9lambda+7` `rArr (4+4lambda)+(2+5lambda)+(9lambda-9)=9lambda+7` `rArr -3+18lambda=9lambda+7` `rArr 9lambda = 10 lambda = 10/9` `:.` Equation of plane `[vecr.(2hati+2hatj-3hatk)-7]+10/9[vecr.(2hati+5hatj+3hatk)-9]=0` `rArr vecr .(18hati+18hatj-27hatk+20hati+50hatj+30hatk)-63-90=0` `rArr vecr.(38hati+68hatj+3hatk) = 153` which is the required equation of the plane.