Ok. In order to compute the eigenvalues, you set the following determinant,

$\displaystyle |A-\lambda I|=\left|\begin{matrix}17-\lambda &16 &16\\ 16 &17-\lambda &16\\ 16 &16 &17-\lambda\end{matrix}\right|,$

equal to zero. The equation you get is $\displaystyle -(\lambda-49)(\lambda-1)^{2}=0.$

The power of each of these factors is its algebraic multiplicity. Hence, $\displaystyle \lambda-49$ is raised to the power 1, and therefore, by definition, the eigenvalue $\displaystyle \lambda=49$ has algebraic multiplicity 1. The factor $\displaystyle \lambda-1$ is raised to the power 2, and therefore, by definition, the eigenvalue $\displaystyle \lambda=1$ has algebraic multiplicity 2.

The geometric multiplicity of an eigenvalue is the dimension of the corresponding eigenspace. When you computed the eigenvectors for $\displaystyle \lambda=1,$ you found two linearly independent eigenvectors. Hence, the geometric multiplicity of the eigenvalue $\displaystyle \lambda=1$ is 2. I have a very strong feeling that when you find the eigenvector (which you should do without delay!) corresponding to $\displaystyle \lambda=49,$ you will find only one eigenvector. Hence, the geometric multiplicity of the eigenvalue $\displaystyle \lambda=49$ is 1.

These are definitions.

Now, I really think you have enough information to solve this problem completely. Keep your head on the target, which is orthogonally diagonalizing $\displaystyle M_{6}.$