The image of the line x−13=y−31=z−4−5 in the plane 2x−y+z+3=0 is the line

3(2)+1(−1)+(−5)(1)=0Given line and given plane are parallel∴ Image line is also parallel to the given lineImage of A (1,3,4) w.r.to given plane lies on the image line.Equation of the normal to the plane is x−12=y−3−1=z−41Any point on the line B=(2r+1,−r+3,r+4)If B is the image of A (1,3,4) then mid point of AB lies on the planeMid point =2r+22,−r+62,r+82Mid point lies in the given plane ⇒22r+22−−r+62+r+82+3=0⇒r=−2,   B(−3,5,2)Image lines is x+33=y−51=z−2−5