$\def\d{\mathrm{d}}\def\Vol{\mathop{\mathrm{Vol}}}$Okay, this really seems not that straigntforward to generalize the previous answer here. However, here is the method which seems to work (and which seemingly is generalizable). This is more or less the same computation written in a more appropriate language.
Let $X$ be the matrix with columns $\mathbf x$, $\mathbf y$ and $\mathbf z$ (writing $X=(\mathbf x,\mathbf y,\mathbf z)$); then $M=XX^\top$. Denote by $\Gamma=X^\top X$ the Gram matrix of the vectors $\mathbf x$, $\mathbf y$, and $\mathbf z$.
It is known that, for any $n\times k$ matrix $A$ and any $k\times n$ matrix $B$ with $n\geq k$, the characteristic polynomials of $AB$ and $BA$ are related by $\chi_{AB}(x)=x^{n-k}\chi_{BA}(x)$, where $\chi_A(x)=\det(xI-A)$. Therefore, in our situation we have
$$
\chi_M(x)=x^2\chi_\Gamma(x).
$$
Next, one can see that
$$
\frac{\d^2}{\d x^2}\chi_M(x)=2\sum_{i<j} \chi_{M_{ij}}(x).
$$
To see this, in the left-hand side, consider the summands where the two operators $\d/\d x$ are applied to the fixed linear terms on the diagonal of $xI-M$; the sum of all those is exactly $\chi_{M_{ij}}(x)$.
To summarize,
$$
p(x):=\sum_{i<j}\chi_{M_{ij}}(x)=\frac12(x^2\chi_\Gamma(x))''=10x^3-18x^2+\alpha x-\beta,
$$
where $\beta=\det\Gamma=\Vol(\mathbf x,\mathbf y,\mathbf z)^2\leq 1$.
Notice that $x^2\chi_\Gamma(x)$ has only real root; hence $p(x)$ has only real roots $\lambda_1\leq \lambda_2\leq\lambda_3$. Notice that for some $i<j$ we have $\chi_{M_{ij}}(\lambda_1)\geq 0$, hence the smallest eigenvalue of $M_{ij}$ does not exceed $\lambda_1$. So, we are interested in estimating $\lambda_1$.
We have
$$
\lambda_1+\lambda_2+\lambda_3=\frac 95, \quad
\frac1{10}\geq \frac\beta{10}=\lambda_1\lambda_2\lambda_3\geq \lambda_1^2\left(\frac95-2\lambda_1\right),
$$
so $20\lambda_1^3-18\lambda_1^2+1\geq 0$. Since $\lambda_1\leq 3/5$, we get that $\lambda_1$ does not exceed the second largest root of $20x^3-18x^2+1$, which is approx. $0.285$. In particular, $\lambda_1\leq 9/20$ with a large gap.
