I'm not sure how much to give away. The matrix being introduced is $ M \; = \; \left( \begin{array}{rrr} 1 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 1 \end{array} \right) , $
which is symmetric, and an orthogonal basis is indeed pretty easy to find. If required, one may then normalize to get an orthonormal basis. If we take a column vector $x$ and its transpose $x^T,$ then the value of the quadratic form at the point $x$ is just $ f(x) = x^T \; M \; x. $ The inner product of two vectors $x,y$ is given by $ \langle x,y \rangle = \frac{1}{2} \left(f(x+y) - f(x) - f(y)\right) = x^T M y = y^T M x$ which is one way to write polarization.
Here is something not mentioned. If you take all three entries in $\vec{x}$ to be integers, the result $f(\vec{x})$ is a nonnegative integer. Not all positive integers show up, though. A number $n \geq 0$ has a "representation" $n=f(x)$ if and only if $n$ is not of the form $4^k (16 m + 14).$ Go figure.