Pg.248 of "Textbook in Tensor Calculus and Differential Geometry" by Prasun Nayak.
Let us suppose that $\lambda_{h|}^i$ is not a unit vector and therefore, the mean curvature $M_h$ in this case is given by
$$M_h=-\frac{R_{ij}\lambda_{h|}^i\lambda_{h|}^j}{g_{ij}\lambda^i_{h|}\lambda^j_{h|}} \tag{1}$$
A proof was not given so I am trying to derive this.
For when $\lambda_{h|}^i$ are components of an orthornormal set of vectors $$M_h=\sum_{k}K_{hk}=\sum_k\frac{\lambda_{h|}^l\lambda_{k|}^i\lambda_{h|}^j\lambda_{k|}^rR_{lijr}}{\lambda_{h|}^l\lambda_{k|}^i\lambda_{h|}^j\lambda_{k|}^r(g_{lj}g_{ir}-g_{lr}g_{ij})}=\sum_k\frac{\lambda_{h|}^l\lambda_{k|}^i\lambda_{h|}^j\lambda_{k|}^rR_{lijr}}{(1)\cdot(1)-0}$$ $$=\lambda_{h|}^l\lambda^j_{h|}R_{lijr}g^{jr}=-\lambda_{h|}^l\lambda^j_{h|}R_{lj}$$
where $\sum_k\lambda^i_{k|}\lambda^r_{k|}=g^{ir}$.
Now for when $\lambda_{h|}^i$ are not components of an orthonormal set of vectors, succumbing to the same approach gives
$$M_h=\sum_k K_{hk}=\frac{\lambda_{h|}^l\lambda_{h|}^j R_{lijr}}{(g_{lj}\lambda_{h|}^l\lambda_{h|}^j)g_{pq}}\sum_k\frac{\lambda_{k|}^i\lambda_{k|}^r}{\lambda_{k|}^p\lambda_{k|}^q} \tag{2}$$ It is not obvious to me how $(2) \to (1)$. Modifying the definition of $M_h$ $$M_h=\frac{\sum_k\lambda_{h|}^l\lambda_{k|}^i\lambda_{h|}^j\lambda_{k|}^rR_{lijr}}{(g_{lj}\lambda_{h|}^l\lambda_{h|}^j)\sum_k g_{pq}\lambda^p_{k|}\lambda_{k|}^q}=-\frac{R_{ij}\lambda_{h|}^i\lambda_{h|}^j}{g_{ij}\lambda^i_{h|}\lambda^j_{h|}N} \tag{3}$$
which is the nearest I can get to $(1)$, but with an extra factor $\frac{1}{N}$.
What's gone wrong? Is there a way to derive $(1)$ or is $(1)$ a definition itself?