If x,y,z are real and $4x^2+9y^2+16z^2-6xy-12yz-8zx=0$ then x,y,z are

Multiply the given expression by 2 and write it as

$(2x-3y{)}^2+(3y-4z{)}^2+(4z-2x{)}^2=0$

$\Rightarrow 2x=3y=4z\Rightarrow \frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in A.P.