We can solve it by star and bar method.
Suppose $1$st person's score is $x_1.$
Suppose $2$nd person's score is $x_2.$
Suppose $3$rd person's score is $x_3.$
We want $x_1+x_2+x_3 = 11$ with constraints $1\leq x_1\leq 6$ , $1\leq x_2\leq 6$ , $1\leq x_2\leq 6$
So $x_1,x_2,x_3$ have minimum values $1$ and remaining $8$ can be scored as $x_1+x_2+x_3 = 8.$
Total no. of ways to score $8=\binom{3+8-1}{2} = \binom{10}{2}.$
Now we have to subtract the ways in which either $x_1$ or $x_2$ or $x_3$ have values $\geq 7 .$
When $x_1 \geq 7,$ we get $x_1+x_2+x_3=2$ (we subtract $6$ from RHS because $x_1$ is already assumed to be $\geq 1$ and $6+1=7$).
No. of ways $= \binom{3+2-1}{2} = \binom{4}{2}$
Doing same for $x_2$ and $x_3.$
Note: Only one of the $3$ can have score $\geq 7$ as if 2 suppose $x_1$ and $x_2$ both is $7$ then total will exceed $8$.
So, total no. of ways = $\binom{10}{2} - \left [ \binom{4}{2} +\binom{4}{2} + \binom{4}{2} \right ] = 27.$
(B) is the correct answer.
Another way -
Contraints - $1\leq x_1,x_2,x_3\leq 6.$
Now, 11 can be broken as -
$\{6,4,1\} \rightarrow$ can be scored in $3! \ ways$
$\{6,3,2\} \rightarrow$ can be scored in $3! \ ways$
$\{5,5,1\} \rightarrow$ can be scored in $\frac{3!}{2!} \ ways$
$\{5,4,2\} \rightarrow$ can be scored in $3! \ ways$
$\{5,3,3\} \rightarrow$ can be scored in $\frac{3!}{2!} \ ways$
$\{4,4,3\} \rightarrow$ can be scored in $\frac{3!}{2!} \ ways$
$3! + 3! +\frac{3!}{2!} + 3! + \frac{3!}{2!} + \frac{3!}{2!}$
$\Rightarrow 3.3! + 3. \frac{3!}{2!} = 18+ 9 =27$