J is just the projector onto the subspace orthogonal to $1_n$. So, consider what A, B and C look like in an (orthonormal) basis in which $1_n$ is the first basis vector. In block form:
$ A = \left[ \begin{array}{cc} a & \vec{b}^T \\ \vec{b} & M \end{array} \right], \quad C = \left[ \begin{array}{cc} 0 & \vec{b}^T \\ 0 & M \end{array} \right], \quad B = \left[ \begin{array}{cc} 0 & 0 \\ 0 & M \end{array} \right] $
As long as M doesn't have any zero eigenvalues then B and C have identical spectra. One sees this as follows. Because A is symmetric, M is symmetric. Let $\vec{v}_1,\ldots,\vec{v}_{n-1}$ be the eigenvectors of M, with corresponding eigenvalues $\lambda_1, \ldots, \lambda_{n-1}$. Then C has the following n-1 right-eigenvectors (in block form):
$ \left[ \begin{array}{c} \lambda_j^{-1} \vec{b} \cdot \vec{v}_j \\ \vec{v}_j \end{array} \right] \quad j=1,\ldots,n-1 $
with corresponding eigenvalues $\lambda_1, \ldots, \lambda_{n-1}$. The final right-eigenvector of C is
$ \left[ \begin{array}{c} 1 \\ \vec{0} \end{array} \right] $ with eigenvalue 0. (Here, $\vec{0}$ indicates the dimension n-1 zero vector.) Similarly, B has the right-eigenvectors
$ \left[ \begin{array}{cc} 0 \\ \vec{v}_j \end{array} \right] \quad j=1,\ldots,n-1 $ with eigenvalues $\lambda_1,\ldots,\lambda_{n-1}$ and
$ \left[ \begin{array}{c} 1 \\ \vec{0} \end{array} \right] $ with eigenvalue 0. I'm not sure what happens when M has one or more zero eigenvalues.