Given $\left|\begin{array}{lll}x & \alpha & 1 \\ \beta & x & 1 \\ \beta & \gamma & 1\end{array}\right|=0$
$\Rightarrow x(x-\gamma)-\alpha(0)+1(\gamma \beta-x \beta)=0$
$\Rightarrow x^{2}-x \gamma+\gamma \beta-x \beta=0$
$\Rightarrow x^{2}-(\gamma+\beta) x+\gamma \beta=0$
$\Rightarrow x=\frac{(\gamma+\beta) \pm \sqrt{\gamma^{2}+\beta^{2}-2 \gamma \beta}}{2}$
( $\because$ roots of Quad. eqn. $a x^{2}+b x+c=0$ are
$\left.\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\right)$
$\Rightarrow x=\frac{(\gamma+\beta) \pm(\gamma-\beta)}{2}$
$\Rightarrow x=\frac{\gamma+\beta+\gamma-\beta}{2}, \frac{\gamma+\beta-\gamma+\beta}{2}$
$\Rightarrow x=\gamma, \beta .$
Hence, roots of the equation $\left|\begin{array}{lll}x & \alpha & 1 \\ \beta & x & 1 \\ \beta & \gamma & 1\end{array}\right|=0$ are
independent of $\alpha$.