Answer:
Step-by-step explanation:
The plane containing line of intersection of two planes x+y-z-2=0 and 4x-y+5z-3 =0
will be of the form
x+y-z-2+k(4x-y+5z-3) =0
Now given that the plane passes through (-1,1,2)
Substitute to get
-1+1-2-2 +k(-4-1+10-3) =0
-4+2k =0
Or k =2
Substitute in the equation of the plane for k to get real equation.
x+y-z-2+2(4x-y+5z-3) =0
Simplify to get
9x-y+9z-8 =0 is the equation of the required plane.