presumably "positively oriented ON-systems" means we are using the "right-hand rule" to determine an orientation (order) of the basis {e_{1},e_{2},e_{3}} (the xyz-axes).
(i am guessing ON stands for orthonormal).
reflection across the plane x = z is definitely NOT the same is "projection on the line x = z" the reflection is an invertible linear map, the projection is NOT.
what i believe he means by the plane "x = z" is the 2-dimensional subspace of R^{3} spanned by {(1,0,1),(0,1,0)}.
let's work out what this reflection does to the vectors (1,0,0), (0,1,0), and (0,0,1). clearly it does nothing to (0,1,0) since that is already in the plane.
suppose v is in the orthogonal complement of P = span({(1,0,1),(0,1,0)}). then our reflection takes v to -v. let's determine P^{⊥}.
if (x,y,z).(1,0,1) = 0, then x = -z.
if (x,y,z).(0,1,0) = 0, then y = 0. so the orthogonal complement is spanned by (1,0,-1).
so let's write (1,0,0) in the basis {(1,0,1),(0,1,0),(1,0,-1)}. this gives:
(1,0,0) = a(1,0,1) + b(0,1,0) + c(1,0,-1) = (a+c,b,a-c) giving:
a+c = 1
b = 0
a-c = 0
thus a = 1/2, b = 0, c = 1/2.
since (1,0,0) = (1/2)(1,0,1) + (1/2)(1,0,-1), applying our reflection to (1,0,0) we see it fixes (1,0,1) and sends (1,0,-1) to (-1,0,1).
thus (1,0,0) gets mapped to (1/2)(1,0,1) - (1/2)(1,0,-1) = (0,0,1).
similarly, if:
(0,0,1) = a(1,0,1) + b(0,1,0) + c(1,0,-1) = (a+c,b,a-c)
a+c = 0
b = 0
a-c = 1
so a = 1/2, b = 0, c = -1/2.
thus (0,0,1) = (1/2)(1,0,1) - (1/2)(1,0,-1) gets mapped to (1/2)(1,0,1) + (1/2)(1,0,-1) = (1,0,0).
so the matrix for our reflection is:
.
now we need to determine the matrix for the linear transformation L(u) = (-1,2,1) x u. so we compute L((1,0,0)), L((0,1,0)) and L((0,0,1)).
L((1,0,0)) = (-1,2,1) x (1,0,0) = (2*0-1*0, 1*1-(-1)*0, (-1)*0-2*1) = (0,1,-2)
L((0,1,0)) = (-1,2,1) x (0,1,0) = (2*0-1*1, 1*0-(-1)*0, (-1)*1-2*0) = (-1,0,-1)
L((0,0,1)) = (-1,2,1) x (0,0,1) = (2*1-1*0, 1*0-(-1)*1, (-1)*0-2*0) = (2,1,0)
so the matrix for L is:
(note this matrix is anti-symmetric, as we would expect since the cross product is anti-commutative).
if we call the matrix for our reflection M, and the matrix for L, N, then the matrix for T is given by MN.