Find the equation of the plane through the intersection of the planes `3x-4y+5z=10a n d2x+2y-3z=4` and parallel to the line `x=2y=3zdot`

Answer

`x-20y+27z=14`

Explanation

Let the reqiuired plane be
`(3x-4y+5z-10)+lambda(2x+2y-3z-4)=0`
`rArr (3+2lambda)x+(2lambda-4)y+(5-3lambda)z-10-4lambda=0`. ……………(i) ,brgt D.r's of normal to the plane are `(3+2lambda),(2lambda-4),(5-3lambda)`.
Given line is `x/6=y/3=z/2`.
D.r.'s of the line are 6,3,2.
`therefore` This line is perpendicular to the plane.
`therefore 6(3+2lambda)+3(2lambda-4)+2(5-3lambda)=0 rArr 12lambda=-16 rArr lambda=-4/3`.