Solving equations

Relevant wiki: Completing the Square - Basic

By rearranging the equations,

$\begin{cases} 9a^2-24b+17=0 \\ 16b^2+4c-7=0 \\ 4c^2-6a+1=0 \end{cases}$

$9a^2-24b+17+16b^2+4c-7+4c^2-6a+1=0$

$9a^2-6a+16b^2-24b+4c^2+4c+11=0$

By completing squares,

$(9a^2-6a+1)+(16b^2-24b+9)+(4c^2+4c+1)=0$

$(3a-1)^2+(4b-3)^2+(2c+1)^2=0$

$\because$ A square number is always positive,

$\begin{cases} 3a-1=0 \Rightarrow a=\frac{1}{3} \\ 4b-3=0 \Rightarrow b=\frac{3}{4} \\ 2c+1=0 \Rightarrow c=-\frac{1}{2} \end{cases}$

$\therefore |abc|=\frac{1}{8}=0.125$