An $11\times 11\times 11$ wooden cube is formed by gluing together $11^3$ unit cubes. What is the greatest number of unit cubes that can be seen from a single point?
$\text{(A) 328} \quad \text{(B) 329} \quad \text{(C) 330} \quad \text{(D) 331} \quad \text{(E) 332}$
[asy]import three; unitsize(1cm);size(100); draw((0,0,0)--(0,1,0),red); draw((0,0,1)--(0,0,0)--(1,0,0)--(1,0,1),red); draw((0,0,1)--(1,0,1),red); draw((0,0,1)--(0,1,1)--(0,1,0)--(1,1,0)--(1,0,0),red); draw((1,1,0)--(1,1,1),red); draw((0,1,1)--(1,1,1)--(1,0,1),red); [/asy]
The best angle for cube viewing is centered on the corner, meaning three of the six faces are visible. So, therefore, the answer is just counting the number of cubes on the three faces, which is $3 \times 10^2$ for the inside parts of the faces, plus $3 \times 10$ for the edges, plus $1$ for the single shared cube in the corner, giving a total of $331$ or $\fbox{D}.$