Suppose $(M,g)$ is a complete Riemann manifold. Is it true that for any $A,B,C \in M$ that do not lie on the same geodesic, there exists an $H \in M$ such that $A,B,C,H$ lie in a totally geodesic submanifold of dimension 2 and $AH \perp BC,BH \perp CA,CH \perp AB$, where $XY$ denotes a geodesic (not necessarily unique) containing $X$ and $Y$ ($X,Y$ can be the same point)? What is the case when $M$ is also real analytic? I know it is true for the standard $\Bbb{S}^n,\Bbb{E}^n$ and $\Bbb{H}^n$ geometries, and I seek proof or counterexamples for general cases.
Edit: According to Didier's answer, there exist smooth Riemann manifolds that admit no totally geodesic submanifold other than geodesic lines and itself, so I will only be asking the real analytic case and the case when $\dim M=2$.
The answer is no, for the simple reason that some Riemannian manifolds $(M,g)$ do not own any totally geodesic submanifold of dimension $ 2 \leqslant k \leqslant \dim M-1$. You can find information concerning this fact in this paper, or in this paper.