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

The equation of plane passing through the line of intersection of plan

1.	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
Answer» Equation of the plane passing through the line of intersection of planes `(2x-y+z)-3=0` and `4x-3y+5z+9=0` is `(2x-y+z-3)+lambda (4x-3y+5z+9)=0" …(i)"` `(2+4lambda)x-(1+3lambda)y+(1+5lambda)z-3+9lambda=0` Since this plane is parallel to the line `(x+1)/(2)=(y+3)/(4)=(z-3)/(5)` whose direction ratio are 2, 4, 5. `therefore `The normal to the plane is perpendicualr to this line. `therefore (2+4lambda)2+[-(1+3lambda)]4+(1+5lamba)5=0` `rArr 4+8lambda-4-12lambda+5lambda25lambda=0` `rArr" "5+21lambda =0` `rArr" "lambda=(-5)/(21)` Substituting `lambda =(-5)/(21)` in equation (i), we get `(2x-y+z-3)+((-5)/(21))(4x-3y+5z+9)=0` `l42x-21y+21z-63-20x+15y-25z-45=0` `22x-6y-4z-108=0` `2(11x-3y-2z-54)=0` `11x-3y-2z-54=0`