1. Find an equation of the plane that passes through the line of intersection of the planes $x-z=1$ and $y+2 z=$ 3, is perpendicular to the plane $x+y=0 z=2$

Step 1 ：\(n_1 = \langle 1, 0, -1 \rangle\) and \(n_2 = \langle 0, 1, 2 \rangle\) are the normal vectors of the given planes.

Step 2 ：The direction vector of the line of intersection is \(d = n_1 \times n_2 = \langle 1, -3, 1 \rangle\).

Step 3 ：The normal vector of the plane perpendicular to \(x+y=0\) and \(z=2\) is \(n_3 = \langle 1, 1, 0 \rangle\).

Step 4 ：The normal vector of the required plane is \(n = d \times n_3 = \langle 1, 1, 4 \rangle\).

Step 5 ：Let \(P = (x, y, z)\) be a point on the line of intersection of the given planes.

Step 6 ：Then \(x - z = 1\) and \(y + 2z = 3\).

Step 7 ：From the first equation, \(z = x - 1\).

Step 8 ：Substituting into the second equation, we get \(y + 2(x - 1) = 3\), so \(y = -2x + 5\).

Step 9 ：Thus, \(P = (x, -2x + 5, x - 1)\).

Step 10 ：Since \(P\) lies on the required plane, the equation of the plane is \(x(-2x + 5) + (x - 1) + 4(x - 1) = 0\).

Step 11 ：Simplifying, we get \boxed{x + y + 4z - 3 = 0}\).