Answer
$x+y+z=4$
Work Step by Step
The plane passes through the line of intersection of the planes $ x-z=1$ and $y+2z=3$ and is perpendicular to the plane $x+y-2z=1$.
Normal vector are: $n_1=\lt 1,0,-1\gt $ and $n_2=\lt 0,1,2\gt $
$d_1=n_1\times n_2=\lt 1,-2,1\gt$
The normal vector of third plane is: $n_3=\lt 1,1,-2\gt $.
Now, $d_2=d_1\times n_3=\lt 3,3,3\gt $
The general form of the equation of the plane is:
$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
or, $ax+by+cz=ax_0+by_0+cz_0$
Plugging in the values ,we get
$3(x-1)+3(y-3)+3(z-0)=0$
After simplification, we get
$3x-3+3y-9+3z=0$
or,
$x+y+z=4$
