Explanation
According to havel-hakimi theorem,
(1, 1, 1 ,1, 1, 1) is graphic if <1,1,1,1,0> is graphic
(0,1,1,1,1) is graphic if(0,1,1,0) is graphic
(0,0,1,1) is graphic 111(0,0,0) is graphic
Since (0,0,0) is graphic (1,1,1,1,1,1) is also graphic.
(The process is always finding maximum degree and removing it from degree sequence, subtracts 1 from each degree for d times from right to left where d is maximum degree)
(2,2,22,2,2) is graphic if (2,2,2,2-1,2-1) = (2,2,2,1,1) is graphic.
(1,1,2,2,2) is graphic if(1,1,1,1)is graphic.
(1,1,1,1) is graphic 1ff (0,1,1)
(0,1,1) is graphic if (0,0) is graphic.
Since (0,0) is graphic (2,2,2,2,2,2) is also graphic.
Consider option C now.
(3,3,3,1,0,0) → (0,0,1.3,3,3) is graphic if (0.0,0,2,2) is graphic.
Note that before applying the havel-hakimi step degree sequence should be in non-increasing order.
(0,0,0,2,2) is graphic if (0,0,-1,1) is graphic.
Since (0,0,-1,1) is not graphic (3,3,3,1,0,0) is also not graphic.