△=0⇒r3−2r−4=0⇒(r−2)(r2+2r+2)=0
∴r=2 as other factor gives imaginary roots. Putting r=2, the equation reduces to 4x−2y+3z=0,x+2y+2z=0 and x+z=0
Putting x+z=0, they reduce to x−2y=0,2y+z=0 and x+z=0.
Choosing x=k, we get y=k2,z=−k which given the required solutions.