Given , x=cy+bz , y=az+cx , z=bx+ay , where x,y,z are not all zero , prove that `a^2+b^2+c^2+2ab=1`

Given x = cy +bz, y = az + cx, z = bx+ ay where x,y and z are not all zero, then a^2 + b^2 + c^2 + 2abc = ___________

If =cy+bz,y=az+cx,z=x+ay, wherex ,yx=cy+bz,y=az+cx,z=x+ay, wherex ,y are not all zeros,then find the value of a^(2)+b^(2)+c^(2)+2abc

Given a=x/(y-z),b=y/(z-x), and c=z/(x-y), where x,y,z and are not all zero,then the value of ab+bc+ca is 0 b.1 c.-1 none of these

Let a, x, y, z be real numbers satisfying the equations ax+ay=z x+ay=z x+ay=az, where x, y, z are not all zero, then the number of the possible values of a is

Given x=cy+bz,y=az+cx and z=bx+ay, then prove a^(2) +b^(2) +c^(2) +2abc =1.

If planes x- cy - bz =0 , cx-y+az =0 and bx + ay - z=0 pass through a staight line the a^(2) +b^(2) +c^(2)=