x+y-3=0, or x+y=3 is a plane A parallel to the z-axis, n1=<1,1,0>
y+z+5=0, or y+z=-5 is a plane B parallel to the x-axis, n2=<0,1,1>
x+z+2=0, or x+z=-2 is a plane C parallel to the y-axis, n3=<1,0,1>, where n represents the normal vector.
We can find the angle θ between the planes by looking at the angle between their normals. For two vectors, a and b, |a||b|cosθ=a·b. |n1|=|n2|=|n3|=√2.
n1·n2=n1·n3=n2·n3=1, therefore 2cosθ=1, θ=π/3 or 60° in each case, so the planes intersect at 60°.
Let x=t, then y=3-t, z=-5-y=-8+t, the parameterised equation of the line intersection of planes A and B;
x=t, y=3-t, z=-2-t, the parameterised equation of the intersection of A and C;
x=t, z=-2-t, y=-5-z=-3+t, the parameterised equation of the intersection of B and C.