Max number of Linearly Independent eigenvectors

I need help with this problem I came across. I give it in a simplified way:

The questions asks for the maximum number of linearly independent eigenvectors for the eigenvalue 7.

Now suppose the Matrix we are dealing with is in 3D and has eigenvectors:

$\displaystyle \{ e_{k_1} \otimes e_{k_2} \otimes e_{k_3} \}$, the k's are natural numbers (including zero).

Now the corresponding eigenvalues are:

$\displaystyle (2k_1+1)+(2k_2+1)+(2k_3+1)$.

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What I thought at first was that since the expression $\displaystyle c_0 \mid e_0 \otimes e_0 \otimes e_0 \rangle $ has eigenvalue 3, adding one more term would mean adding something of the form $\displaystyle c_1 \mid e_1 \otimes e_0 \otimes e_0 \rangle $, but the added expression would have eigenvalue 8. So at most we can have 1 eigenvector.

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I'm not sure if just adding eigenvalues in the way I did is allowed.

UPDATE:

I was thinking maybe the question meant the following:

Given the eigenvalue $\displaystyle (2k_1+1)+(2k_2+1)+(2k_3+1)$, if we equate this to seven, then we obtain $\displaystyle k_1+k_2+k_3=2$, in which case there are a maximum of 6 linearly independent eigevectors:

$\displaystyle \mid e_1 \otimes e_1 \otimes e_0 \rangle $ , $\displaystyle \mid e_0 \otimes e_1 \otimes e_1 \rangle $ , $\displaystyle \mid e_1 \otimes e_0 \otimes e_1 \rangle $

$\displaystyle \mid e_2 \otimes e_0 \otimes e_0 \rangle $ , $\displaystyle \mid e_0 \otimes e_2 \otimes e_0 \rangle $ , $\displaystyle \mid e_0 \otimes e_0 \otimes e_2 \rangle $

Which is Right?

-Thanks!