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

Let the required equation of the plane be
`(2x+y-z-3)+lambda(5x-3y+4z+9)=0`
`rArr (2+5lambda)x+(1-3lambda)y+(-1+4lambda)z-3+9lambda=0`………….(i)
The plane given by (i) is parallel to the line
`(x-1)/2=(y-3)/4=(z-5)/5`……………….(ii)
So, the normal to the plane (i) is perpendicular to the line (ii).
`therefore 2(2+5lambda)+4(1-3lambda)+5(-1+4lambda)=0`
`rArr (4+10lambda)+(4-12lambda)=0`
`rArr 18lambda=-3 rArr lambda=-3/18 rArr lambda=-1/6`.
Putting `lambda=-1/6` in (i), we get
`(2-5/6)x+(1+3/6)y+(-1-4/6z)-9 xx 1/6=0`.
`rArr (7x)/6+(9y)/6-(10z)/6-27/6=0`
`rArr 7x+9y-10z-27=0`.
Hence, the required equation of the plane is `7x+9y-10z-27=0`.