Given that n belongs to T if $n^{2}+3n+2$ is divisible by 6.
So we can say that n belongs to T if $(n^{2}+3n+2)$ $mod$ $6$$=0$
but $n^{2}+3n+2=(n+1)(n+2)$.
So $(n^{2}+3n+2)mod 6=((n+1)(n+2))mod6$
$\Rightarrow$$((n+1)(n+2))mod6=((n+1)mod6*(n+2)mod6)mod6$
$(n^{2}+3n+2)mod 6=((n mod 6+1)(n mod 6 +2))mod 6$
let us check some elements from set S.
for n=0,$((n mod 6+1)(n mod 6 +2))mod 6$$=2$$\neq$0
Therefore,0 does not belong to T.
$\therefore 0+6,0+2*6,0+3*6,0+4*6$ does not belong to T
i.e.,$6,12,18,24$ doesn't belong to T.
for n=1,$((n mod 6+1)(n mod 6 +2))mod 6$$=0$
$\therefore 1$ belongs to T.
$\therefore 1,1+1*6,1+2*6,1+3*6,1+4*6$ belongs to T
i.e$1,7,13,19,25$ belongs to T.
for n=2,$((n mod 6+1)(n mod 6 +2))mod 6$$=0$.
$\therefore 2$ belongs to T.
$\therefore 2,2+1*6,2+2*6,2+3*6$ belongs to T
i.e$2,8,14,20$ belongs to T.
for n=3,$((n mod 6+1)(n mod 6 +2))mod 6$$=2$$\neq$0
Therefore,0 does not belong to T.
$\therefore 3,3+1*6,3+2*6,3+3*6$ does not belong to T
i.e.,$3,9,15,21$ doesn't belong to T.
for n=4,$((n mod 6+1)(n mod 6 +2))mod 6$$=0$.
$\therefore 4$ belongs to T.
$\therefore 4,4+1*6,4+2*6,4+3*6$ belongs to T
i.e$4,10,16,22$ belongs to T.
for n=5,$((n mod 6+1)(n mod 6 +2))mod 6$$=0$.
$\therefore 5$ belongs to T.
$\therefore 5,5+1*6,5+2*6,5+3*6$ belongs to T
i.e$5,11,17,23$ belongs to T.
Here we need not check further because we have indirectly checked all the other elements(i.e., 7-25) while checking 0 to 6..
$\therefore$$T=${$1,2,4,5,7,8,10,11,13,14,16,17,19,20,22,23,25$}
or
$T=S-${$0,3,6,12,18,24,9,15,21$}
Hence it's cardinality =17
Hence option B is the answer.
Here actually,$(n^{2}+3n+2)$mod 6 is =((n mod 6+1)mod 6 *(n mod 6 +2)mod 6)mod 6.
But i have written it as $(n^{2}+3n+2)$mod 6 =((n mod 6+1)(n mod 6 +2))mod 6$.
This is because, the modulo 6 (in the bold) just complicates the evaluation of expression.. and it's removal won't affect the answer..