$x^2+4y^2+16z^2=48$$xy+4yz+2zx=24$$\{x,y,z\}\equiv$

(A) $\{  \mp 4, \pm 2, \pm 1\} $
(B) $\{  \pm 4, \mp 2, \pm 1\} $
(C) $\{  \pm 4, \pm 2, \mp 1\} $

(D) $\{  \pm 4, \pm 2, \pm 1\} $

Solution

$(x+2y+4z)^2=x^2+4y^2+16z^2+4xy+8zx+16yz$

$=x^2+4y^2+16z^2+4(xy+2zx+4yz)$

$=48+4\times 24 =144$

$\Rightarrow x+2y+4z=\pm 12$ ......(*)

Now, $x^2+4y^2+16z^2=48=2\times 24 =2(xy+4yz+2zx)$

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

$\Rightarrow \frac{1}{2}\left[ {{{(x - 2y)}^2} + {{(2y - 4z)}^2} + {{(4z - x)}^2}} \right] = 0$

$\Rightarrow x=2y=4z$

From (*), $x+x+x=\pm 12$

$\Rightarrow x=\pm 4$, $y=\pm 2$, $z=\pm 1$

Answer: Option (D)