If it is only one quadratic equation $x^2-xy+y^2=a^2$, we can get some integral solutions as follows. \begin{align*} &\left\{ \begin{split} x&=k(2mn-n^2)\\ y&=k(m^2-n^2)\\ a&=k(m^2-mn+n^2)\\ \end{split}\right. &\quad \left\{ \begin{split} x&=k(m^2-2mn)\\ y&=k(2mn-n^2)\\ a&=k(m^2-mn+n^2)\\ \end{split}\right.\\ \\ &\qquad\qquad\downarrow p=m-n&p=m-n\downarrow\qquad\qquad\qquad\\ \\ &\left\{ \begin{split} x&=k(2np+n^2)\\ y&=k(2np+p^2)\\ a&=k(p^2+np+n^2)\\ \end{split}\right. &\quad \left\{ \begin{split} x&=k(p^2-n^2)\\ y&=k(2np+n^2)\\ a&=k(p^2+np+n^2)\\ \end{split}\right.\\ \end{align*}
However, I'm not sure if these are the complete solutions of Diophantine equation $x^2-xy+y^2=a^2$.
But how to solve this Diophantine equation system in integers? \begin{align*} \left\{ \begin{split} \large{x^2-xy+y^2}&\large{=a^2}\\ \large{y^2-yz+z^2}&\large{=b^2}\\ \large{x^2-xz+z^2}&\large{=c^2}\\ \end{split}\right. \end{align*} I got some non-trivial examples:
$ \begin{align*} \left\{ \begin{split} x&=\phantom{0}7\\ y&=15\\ z&=40\\ a&=13\\ b&=35\\ c&=37 \end{split}\right. \end{align*}$,$\begin{align*} \left\{ \begin{split} x&=\phantom{0}21\\ y&=\phantom{0}56\\ z&=120\\ a&=\phantom{0}49\\ b&=104\\ c&=111 \end{split}\right. \end{align*}$,$\begin{align*} \left\{ \begin{split} x&=\phantom{0}77\\ y&=117\\ z&=165\\ a&=103\\ b&=147\\ c&=143 \end{split}\right. \end{align*}$.
https://benvitalenum3ers.wordpress.com/2016/12/20/make-x2-xy-y2-x2-xz-z2-y2-yz-z2-squares-part-4/
Dickson book (history of theory of numbers) volume 2 page 511 has solution.
$x=(n^2-1)(m^2-1)$
$y=(2n-1)(m^2-1)$
$z=(n^2-1)(2m-1)$
Where, $m=2×(2q^2-pq-qv)/(3q^2-2pv+pq-2p^2)$
$(p,q,v)=((2n-1),(n^2-1),(n^2-n+1))$
For n=3, $(p,q,r)=(5,8,7)$
& $m=(4/7)$
$(x,y,z)=((-264),(-165),(56))$
$(a,b,c)=(231,199,296)$