Let x∈R and let
[1 1 1] [2 x x]
P = [0 2 2] Q = [0 4 0]
[0 0 3] [x x 6[

and R = PQP⁻¹. Then which of the following options is/are correct?

A. There exists a real number x such that PQ = QP
B. det R = det [2 x x] + 8, for all x ∊ R
[0 4 0]
[x x 5]
C. For x = 0, if R [1] = 6 [1], then a + b = 5
[a] [a]
[b] [b]
D. For x = 1, there exists a unit vector αi + βj + γk for which R [α] = [0]
[β] [0]
[γ] [0]