Here's my setup and a numeric solution to the problem; I'm not sure how to find an explicit solution as of now, I'll keep researching.
Anyway, first note that: $\begin{align}A^2 &= a_1^2x^2+2a_1a_2xy + 2a_1a_3xz+a_2^2y^2 +2a_2a_3yz+a_3^2z \\ B^2 &= b_1^2x^2+2b_1b_2xy + 2b_1b_3xz+b_2^2y^2 +2b_2b_3yz+b_3^2z \\C^2 &= c_1^2x^2+2c_1c_2xy + 2c_1c_3xz+c_2^2y^2 +2c_2c_3yz+c_3^2z\end{align}$
Thus, after some factoring,
$A^2+B^2+C^2 = (a_1^2+b_1^2+c_1^2)x^2 + 2(a_1a_2+b_1b_2+c_1c_2)xy + 2(a_1a_3+b_1b_3+c_1c_3)xz$ $+ (a_2^2+b_2^2+c_2^2)y^2 + 2(a_2a_3+b_2b_3+c_2c_3)yz+(a_3^2+b_3^2+c_3^2)z^2$
It follows that:
$\alpha A^2 + \beta B^2 + \gamma C^2 = (\alpha a_1^2+\beta b_1^2+\gamma c_1^2)x^2 + 2(\alpha a_1a_2+\beta b_1b_2+\gamma c_1c_2)xy + 2(\alpha a_1a_3+\beta b_1b_3+\gamma c_1c_3)xz$ $+ (\alpha a_2^2+\beta b_2^2+\gamma c_2^2)y^2 + 2(\alpha a_2a_3+\beta b_2b_3+\gamma c_2c_3)yz+(\alpha a_3^2+\beta b_3^2+\gamma c_3^2)z^2$
As
$2x^2+3y^2-2yz+3z^2 = $
$(a_1^2+b_1^2+c_1^2)x^2+ 2(a_1a_2+b_1b_2+c_1c_2)xy + 2(a_1a_3+b_1b_3+c_1c_3)xz$ $+ (a_2^2+b_2^2+c_2^2)y^2 + 2(a_2a_3+b_2b_3+c_2c_3)yz+(a_3^2+b_3^2+c_3^2)z^2$
and
$x^2+6xy+3y^2+2yz-6zx+3z^2 = $
$(\alpha a_1^2+\beta b_1^2+\gamma c_1^2)x^2 + 2(\alpha a_1a_2+\beta b_1b_2+\gamma c_1c_2)xy + 2(\alpha a_1a_3+\beta b_1b_3+\gamma c_1c_3)xz$ $+ (\alpha a_2^2+\beta b_2^2+\gamma c_2^2)y^2 + 2(\alpha a_2a_3+\beta b_2b_3+\gamma c_2c_3)yz+(\alpha a_3^2+\beta b_3^2+\gamma c_3^2)z^2$
The sums of the $a_i, b_i, c_i$ must be equal to the coefficients on the polynomials in question. Thus, we can build a system of nonlinear equations on the $a_i, b_i, c_i$ and $\alpha,\beta,\gamma$:
$\begin{align}2&=a_1^2+b_1^2+c_1^2 \\3&= a_2^2+b_2^2+c_2^2 \\ 3&=a_3^2+b_3^2+c_3^2 \\ 0&=a_1a_2+b_1b_2+c_1c_2 \\ 0&=a_1a_3+b_1b_3+c_1c_3 \\ -1&=a_2a_3+b_2b_3+c_2c_3 \\ 1&=\alpha a_1^2+\beta b_1^2+\gamma c_1^2 \\ 3&=\alpha a_2^2+\beta b_2^2+\gamma c_2^2 \\ 3&=\alpha a_3^2+\beta b_3^2+\gamma c_3^2 \\ 1&=\alpha a_2a_3+\beta b_2b_3+\gamma c_2c_3 \\ 3&=\alpha a_1a_2+\beta b_1b_2+\gamma c_1c_2 \\ 3&=\alpha a_1a_2+\beta b_1b_2+\gamma c_1c_2\end{align}$
Which is a system of 12 equations in 12 variables. Unfortunately, as it's nonlinear, an explicit solution is difficult to find. Fortunately, Maple happily provides a numeric solution: $\alpha = 2 , \beta = -1 , \gamma = 2,$
$ a_1 = .9596238496 , a_2 = 1.240910300 , a_3 = -0.6783373986 ,$
$ b_1 = -1 , b_2 = 1, b_3= -1 ,$
$ c_1 = -0,2812864509 , c_2 = 0.6783383986, c_3 = 1.240910300$
Which are accurate to $10^{-9}$ Unfortunately, as this is only a numerical solution, I have no way to verify the nature of the solution set