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

Explanation

The equation of the plane passing through the line of intersection of the planes `2x - y + z = 3` and `4x - 3y + 5z + 9 = 0` is
`(2x - y + z - 3) + lambda (4x - 3y + 5z + 9) = 0`
i.e., `(2+4lambda) x + (-1-3lambda) + (1+5lambda) z + (-3 + 9lambda) = 0`
The direction ratios of the normal to this plane are `2 + 4lambda, -1 -3lambda, 1 + 5lambda.`
Since this plane is parallel to the line
` (x+1)/(2) = (y + 3)/(4) = (z-3)/(5)` whose direction ratios are 2, 4, 5, the normal to the plane is perpendicular to this line.
`therefore 2(2+4lambda) + 4(-1-3lambda) + 5(1+5lambda) = 0`
`therefore 4 + 8 lambda -4- 12lambda + 5 + 25 lambda =0`
` therefore 21 lambda + 5 = 0 " " therefore lambda = (-5)/(21)`
Substituting `lambda = -(5)/(21)` in equation (1), we get,
`(2x -y + z - 3) - (5)/(21) (4x -3y + 5z + 9) =0`
i.e., `42x - 21y +21 z - 63 - 20x + 15y - 25z - 45 = 0`
i.e., `22x - 6y - 4z = 108`
i.e., `11x - 3y - 2z = 54`
This is the required equation of the plane.