Answer
Not Reflexive
Symmetric
Not Transitive
Work Step by Step
Reflexive : for all x $\in$ Z , xOx <=> x-x is odd
Counter Example : take any int lets say 2, 2-2 = 0 zā 2k+1 (hence (x,x) $\notin$ O)
Symmetric : for all x,y $\in$ Z , if xOy then yOx
xOy <=> x-y = 2k+1
-x+y = -2k - 1 (Multiply both sides by -1)
y- x = 2{-k} - 1 (Commutive Law)
y-x = odd (note that -k $\in$ Z hence y-x is odd)
yOx (by Def of O)
Transitive: for all x,y and z $\in$ Z if (xOy and yOz) then ( xOz)
Counter Example: take x= 1, y=2 ,z=1
xOy = -1 #odd
yOz = 1 #odd
but xOz = 0 # even , hence the statement is false, hence not transitive)