# If $ax^2_1+by^2_1+cz^2_1=ax^2_2+by^2_2+cz^2_2=ax^2_3+by^2_3+cz^2_3=d$ and $ax_2x_3+by_2y_3+cz_2z_3=ax_3x_1+by_3y_1+cz_3z_1=ax_1x_2+by_1y_2+cz_1z_2=f$,

If $ax^2_1+by^2_1+cz^2_1=ax^2_2+by^2_2+cz^2_2=ax^2_3+by^2_3+cz^2_3=d$ and $ax_2x_3+by_2y_3+cz_2z_3=ax_3x_1+by_3y_1+cz_3z_1=ax_1x_2+by_1y_2+cz_1z_2=f$, then prove that $$\begin{vmatrix} x_1 &y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3\\ \end{vmatrix}=(d-f)\sqrt{\frac{d+2f}{abc}}$$ where $a,b,c\ne0$

$\begin{vmatrix} x_1 &y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3\\ \end{vmatrix}=x_1y_2z_3-x_1y_3z_2-y_1x_2z_3+y_1x_3z_2+z_1x_2y_3-z_1x_3y_2$
I cannot understand how I should use the given equation to convert it in terms of $a,b,c,d,f.$ Please help.

• Are you sure $\left(d-f\right)$ shouldn't be $\pm\left(d-f\right)$ ? Swapping $b, y_1, y_2, y_3$ with $c, z_1, z_2, z_3$ flips the sign of the left-hand side. – darij grinberg Aug 9 '18 at 17:02