If x2+9y2+25z2=xyz15x+5y+3z, then

x2+9y2+25z2=15yz+5zx+3xy⇒(x)2+(3y)2+(5z)2−(x)(3y)−(3y)(5z)−(x)(5z)=0⇒12(x−3y)2+(3y−5z)2+(x−5z)2=0⇒x−3y=0,3y−5z=0,x−5z=0x=3y=5z⇒x:y:z=11:13:15Therefore, 1/x, 1/y, and 1/z are in A.P. and x, y, and z are in H.P.