If $21\left({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2\right)=(\mathrm{x}+2\mathrm{y}+4\mathrm{z}{)}^2$, then $\mathrm{x},\mathrm{y},\mathrm{z}$ are in

$\because 21=1^2+2^2+4^2$

Now, $\left(1^2+2^2+4^2\right)\left({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2\right)=(\mathrm{x}+2\mathrm{y}+4\mathrm{z}{)}^2$

$\Rightarrow \left(1^2+2^2+4^2\right)\left({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2\right)-(1\cdot \mathrm{x}+2\cdot \mathrm{y}+4.\mathrm{z}{)}^2=0$

$\Rightarrow \left|\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\\ \mathrm{x}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{y}\end{array}\right|+\left|\begin{array}{l}2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\\ \mathrm{y}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{z}\end{array}\right|+\left|\begin{array}{l}4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ \mathrm{z}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\end{array}\right|=0$  (by Lagrange’s identity)

$\Rightarrow (\mathrm{y}-2\mathrm{x}{)}^2+(2\mathrm{z}-4\mathrm{y}{)}^2+(4\mathrm{x}-\mathrm{z}{)}^2=0$

Which is possible only when

$y-2x=0,2z-4y=0\text{ and }4x-z=0$ $\text{ or }\frac{\mathrm{y}}{\mathrm{x}}=2,\frac{\mathrm{z}}{\mathrm{y}}=2,\frac{\mathrm{z}}{\mathrm{x}}=4$

$\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{z}}{\mathrm{y}}\Rightarrow {\mathrm{y}}^2=\mathrm{zx}\therefore \mathrm{x},\mathrm{y},\mathrm{z}\text{ are in G.P. }$