Partial answer:

Let $A = \left( \begin{array}{ccc}
a & b & c \\
\overline{b} & e & d \\
\overline{c} & \overline{d} & f \end{array} \right) $ be the generic hermitian matrix with octonionic coefficients. This means that $b,c,d$ can be written as $X_1.1 +X_2.i_1 + \cdots + X_8.i_7$, where the $X_p$ are abstract variables (which will take value in $K$) and $1,i_1, \cdots, i_7$ are basis of $\mathbb{O}$ over $K$.

The symbols $\overline{b}, \overline{c}, \overline{d}$ are the conjugate of $b,c,d$ and $a,e,f$ are self conjugate, that is they are of the form $X_1.1$, with $X_1$ a variable which will take value in the field $K$. Overall, you have $27$ abstract variables $X_1,\cdots X_{27}$ (which will take value in $K$) coming into the picture.

Now, if you compute de determinant of $A$ (using Sarrus rule for instance), you will find a cubic equation in $X_1, \cdots, X_{27}$ which involves only integer numbers (the purely octonionic numbers disappear, which is kind of extraordinary). This computation makes sense because of the pseudo-associativity of $\mathbb{O}$, you don't need it to be commutative or even associatif.

This equation is the equation of a cubic hypersurface in $\mathbb{P}^{26}$ and it is irreducible. Now let $Z$ be the scheme cut out by the $2 \times 2$-minors of $A$. Again one finds that the minors are equations in $X_1, \cdots, X_{27}$ with only integers coefficients (again the purely octonionic coefficients vanish).

Each minor is a quadratic equation in $X_1, \cdots, X_{27}$ and it is easy to see that they correspond to the partial derivatives of the equation of $det A$ with respect to each variables $X_1, \cdots, X_{27}$. Hence, the scheme $Z$ is the singular locus of $det A = 0$.

The scheme $Z$ is smooth. Indeed, if it had a singular point, this point woud be a triple point of $det A =0$. Being a cubic hypersurface, $det A =0$ would then be a cone, which isn't the case (this is a bit more complicated to check).

Now, $Z$ being the singular locus of $det A=0$ and this hypersurface being cubic, Bezout's theorem insures that every line meeting two points of $Z$ will be included in $det A=0$. This proves that $S(Z) \subset \{ det A = 0 \}$, where $S(Z)$ is the secant variety of $Z$.

Now two things are left to prove :

_$\dim Z = 16$,

_ the secant variety $S(Z)$ is actually the whole hypersurface $det A =0$.

I don't remember a simple proof of the first fact. Let's admit it. Since $Z$ is smooth of dimension $16$ and included in $\mathbb{P}^{26}$, ZAK's theorem on tangency gurantees that the dimension of $S(Z)$ must be at least $25$. The hypersurface $det A = 0$ being irreducible, we have $S(Z) = \{det A = 0 \}$.

**EDIT** : There is a mistake in the description of the $2 \times 2$ minors. Three of them will indeed be quadartic equations over $K$ with integer coefficients. The other $3$ will have fully octonionic equations, each of them giving $8$ equations over $K$ with integer coeffcients. At the end, wet $27$ equations over $K$ with integer coefficients, corresponding indeed to the $27$ partial derivatives of the equation of $det A$.