Find the equation of the plane passing through the line of intersection of the planes 2x + y - Z = 3,5x - 3y + 4z + 9 = 0 and parallel to the line `(x-1)/2=(y-3)/4=(z-5)/5`

Explanation

The equation of a plane passing through the intersection of the given planes is gives by
`2x+y-z-3)+lambda(5x-3y+4z+9) =0` for some real number `lambda`
`rarr (2+ 5 lambda)x+(1-3 lamdba)y+(4lambda-1)z+(9lambda-3)=0`
If this plane is parallel to the line `(x-1)/(2)=(y-3)/(5)=(z-5)/(5)` then the normal to the plane is perpendicuale to this line
`rarr (10 lambda -12 lambda + 20 lambda)+(4+4-5)=0`
`rarr 18 lambda=-3 rarr lambda =(-3)/(18)=(-1)/(6)`
putting `lambda =(-1)/(6)` in (i) we get
`(2-(5)/(6))x+(1+(3)/(6))y+(-4)/(6)-1)z+(-9)/(6)-3)=0`
`rarr 7x+9y -10z-27 =0`
Hence the required equation of the plane is 7x+9y-10z-27 =0