Or you can form the alternating sum of the bits, e.g. for 0b10011001 you calculate 1-0+0-1+1-0+0-1 = 0 which is divisible by three. (That's similar to the divisibility test by 11 of a number in base-10, or more generally testing if a number in base `b` is divisible by b+1)
Note that each tromino covers one yellow square, one red square and one green square. Also, there are 21 yellow squares, 21 green squares and 22 red squares, so the square which will be left must be red.
and again there is (obviously) one more red square than yellow squares and green squares. The square which will be left empty, should be red in both of the colorings (the original coloring and its reflection). The only squares that satisfy that, are (3, 3), (6, 3), (3, 6) and (6, 6).
