If a,b,care non-zero, then the number of solutions of 2x2a2−y2b2−z2c2=0; −x2a2+2y2b2−z2c2=0; −x2a2−y2b2+2z2c2=0is

(2)−(3)⇒3y2b2−3z2c2=0⇒y2b2=z2c2,substitute in equation(1)(1)⇒2x2a2−2y2b2=0⇒x2a2=y2b2∴x2a2=y2b2=z2c2=k2(say)x2=a2k2,y2=b2k2,z2=c2k2⇒x=±ka,y=±kb,z=±kc,k∈R.