Đặt `x/3`=`y/4`=`z/5`=k

Từ đó: x=3k, y=4k, z=5k

$2x^2$+$2y^2$+$3z^2$=100

2.$(3k)^2$+2.$(4k)^2$-3.$(5k)^2$=100

-25$k^2$=100

$k^2$=100:(-25)

$k^2$=-4

k=`+-`$\sqrt{4}$=`+-`2

TH1: k=2

$\Rightarrow$x=2.3=6

y=2.4=8

z=2.5=10

TH2: k=-2

$\Rightarrow$x=(-2).3=-6

y=(-2).4=-8

z=(-2).5=-10

Vậy: x=6; y=8; z=10

x=-6; y=-8; z=-10