 Solution

A single eigenbasis works for every choice of the parameters. The columns $v_1=(1,0,0{)}^T$, $v_2=(1,1,0{)}^T$, $v_3=(1,1,1{)}^T$ satisfyThey are linearly independent because their column matrix has determinant one. Thus the diagonalization isThis formula includes all coincidences among $a,b,c$, so no special repeated-eigenvalue cases need separate bases.