Invertibility of sequence of symmetric square matrices with fixed rank

Question

We have $X= \begin{bmatrix} 1 & a & b \\ a & h & c\\ b & c & h \\ \end{bmatrix} $, $X$ (square, symmetric and at least one element of the diagonal is $1$ and $-1<h<1$ ), assume a sequence of matrices $X_n= \begin{bmatrix} 1 & a_n & b_n \\ a_n & h_n & c_n\\ b_n & c_n & h_n \\ \end{bmatrix} $, where Rank(X_n)=Rank(X). Does that implies that $X$ is invertible? if $X$ is not invertible can the generalized inverse of $X$ be $X^{+}= \begin{bmatrix} 0 & g_{12} & g_{13} \\ g_{14} & g_{15} & g_{16}\\ g_{17} & g_{18} & g_{19} \\ \end{bmatrix} $ or the zero element can never happen given the structure of $X$ (square, symmetric and at least one element of the diagonal is 1 and $-1<h<1$)??