If $\mathrm{xcos}\mathrm{\theta }=\mathrm{ycos}\left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)=\mathrm{zcos}\left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)$,then the value of  $\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}$ is equal to 

We have, $\mathrm{xcos}\mathrm{\theta }=\mathrm{ycos}\left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)$$=\mathrm{zcos}\left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)=\mathrm{k}$ 

$\begin{array}{r}\begin{array}{ll}\Rightarrow \cos \mathrm{\theta }=\frac{\mathrm{k}}{\mathrm{x}},\cos \left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)=\frac{\mathrm{k}}{\mathrm{y}}\text{ and }\cos \left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)=\frac{\mathrm{k}}{\mathrm{z}}& \\ \begin{array}{rl}\frac{\mathrm{k}}{\mathrm{x}}+\frac{\mathrm{k}}{\mathrm{y}}+\frac{\mathrm{k}}{\mathrm{z}}& =\cos \mathrm{\theta }+\cos \left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)+\cos \left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)\\ & =\cos \mathrm{\theta }-\cos \left(\frac{\mathrm{\pi }}{3}-\mathrm{\theta }\right)-\cos \left(\frac{\mathrm{\pi }}{3}+\mathrm{\theta }\right)\end{array}& \\ =\cos \mathrm{\theta }-2\cos \frac{\mathrm{\pi }}{3}\cos \mathrm{\theta }=0& \\ \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}=0& \end{array}\end{array}$