ya
the in that case wat i think is
firstly find teh line of intersection of the two planes(that u can find i guess)
now take any point on that line suppose(p,q,r)
now that point will lie on the image plane
assume direction ratios of the normal to the image plane be (a,b,c)
so eq of image plane is a(x-p)+b(y-q)+c(z-r)=0
let teh direction ratios of incident plane be(g,h,j)
now find the angle between the incident plane and plane mirror,,,,let that be α
so angle between the incident plane and image palne wiill be 2α....hence angle betwenn the normals form des 2 planes will be 2α
so use cos(2α)=ga+hb+cr√a2+b2+c2√g2+h2+j2
except(a,b,c) everthinmg is known u hav to find (a,b,c)
now use the condition that the teh line of intersection of the two planes lies on the image plane
if (r1,p1,q1,) be direction ratio of line of intersection
so r1a+p1b+q1c=0
u will get two eqautions in a b c
find it and hence the image plane