If a, b, c are real numbers, then the intervals in which $f\left(x\right)=\left|\begin{array}{l}x+a^{2}abac\\ abx+b^{2}bc\\ acbcx+c^2\end{array}\right|\text{\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}strictly\hspace{0.17em}\hspace{0.17em}decreasing}$

$\begin{array}{r}f\left(x\right)=\frac{1}{abc}\left|\begin{array}{ccc}a\left(x+a^2\right)& a{b}^2& a{c}^2\\ a^{2}b& b\left(x+b^2\right)& b{c}^2\\ a^{2}c& b^{2}c& c\left(x+c^2\right)\end{array}\right|\\ =\frac{abc}{abc}\left|\begin{array}{ccc}x+a^2& b^2& c^2\\ a^2& x+b^2& c^2\\ a^2& b^2& x+c^2\end{array}\right|\\ =\left|\begin{array}{ccc}x+a^2+b^2+c^2& b^2& c^2\\ x+a^2+b^2+c^2& x+b^2& c^2\\ x+a^2+b^2+c^2& b^2& x+c^2\end{array}\right|\\ =\left(x+a^2+b^2+c^2\right)\left|\begin{array}{ccc}1& b^2& c^2\\ 1& x+b^2& c^2\\ 1& b^2& x+c^2\end{array}\right|\end{array}$

$\begin{array}{r}=x^2\left(x+a^2+b^2+c^2\right)\\ f^{\mathrm{\prime }}\left(x\right)=3x^2+2x\left(a^2+b^2+c^2\right)\\ f^{\mathrm{\prime }}\left(x\right)<0\\ x(3x+2)\left(a^2+b^2+c^2\right)\le 0\\ -\frac{2}{3}\left(a^2+b^2+c^2\right)\le x\le 0\end{array}$

$\begin{array}{r}x(3x+2)\le 0\\ -\frac{2}{3}\le x\le 0\end{array}$