$\left|\begin{array}{ccc}{2} & {2 a} & {a} \\ {2} & {3 b} & {b} \\ {2} & {4 c} & {c}\end{array}\right|=0, \Rightarrow\left|\begin{array}{ccc}{1} & {2 a} & {a} \\ {0} & {3 b-2 a} & {b-a} \\ {0} & {4 c-2 a} & {c-a}\end{array}\right|=0$
$\Rightarrow(3 b-2 a)(c-a)-(b-a)(4 c-2 a)=0$
$\Rightarrow 2 \mathrm{ac}=\mathrm{bc}+\mathrm{ab}$
$\Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}$
Hence $\frac{1}{\mathrm{a}}, \frac{1}{\mathrm{b}}, \frac{1}{\mathrm{c}}$ are in A.P.