Let $\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{R}$such that no two of them are equal and satisfy $\left|\begin{array}{ccc}2\mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{b}& \mathrm{c}& 2\mathrm{a}\\ \mathrm{c}& 2\mathrm{a}& \mathrm{b}\end{array}\right|=0$

then equation $24{\mathrm{ax}}^2+4\mathrm{bx}+\mathrm{c}=0$has

Given determinant , 

$6\mathrm{abc}-8{\mathrm{a}}^3-{\mathrm{b}}^3-{\mathrm{c}}^3=0$ (or)
$(2\mathrm{a}+\mathrm{b}+\mathrm{c})\left[(2\mathrm{a}-\mathrm{b}{)}^2+(\mathrm{b}-\mathrm{c}{)}^2+(\mathrm{c}-2\mathrm{a}{)}^2\right]=0$
(or)

$2\mathrm{a}+\mathrm{b}+\mathrm{c}=0$

$\text{ Let }\mathrm{f}\left(\mathrm{x}\right)=8{\mathrm{ax}}^3+2{\mathrm{bx}}^3+\mathrm{cx}\Rightarrow \mathrm{f}\left(0\right)=0$

$$\begin{gathered} and \\ f\left(1\right)=8a+4b+4c \\ =4(2a+b+c) \\ =0 \end{gathered}$$

$$\begin{gathered} \text{ Hence }f\left(0\right)=f\left(1\right) \\ \text{ Then applying Rolle's Theorem, we get } \\ \text{ Hence there exists atleast one root of the given polynomial between }[0,1] \end{gathered}$$
$\Rightarrow \mathrm{f}\left(\frac{1}{2}\right)=\mathrm{a}+\frac{\mathrm{b}}{2}+\frac{\mathrm{c}}{2}=\frac{2\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=0$
So, f(x) satifies the Rolle’s theorem and, hence, f'(x)=0 has at least one root in $\left[0,\frac{1}{2}\right]$