If the system of equations −ax+y+z=0, x−by+z=0  and x+y−cz=0 (a,b,c≠−1)  has a non-zero solution, then 11+a+11+b+11+c=

The given system of linear homogeneous equations has a non-trivial solution ⇒|−a     1       1 1      −b    1  1        1    −c |=0  ⇒−a(bc−1)−1(−c−1)+(1+b)=0  ⇒abc=a+b+c+2  ……. (1)Also, 11+a+11+b+11+c  =(1+b)(1+c)+(1+c)(1+a)+(1+a)(1+b)(1+a)(1+b)(1+c)= 3+2(a+b+c)+(bc+ca+ab)1+(a+b+c)+(bc+ca+ab)+abc = 3+2(a+b+c)+(bc+ca+ab)1+(a+b+c)+(bc+ca+ab)+2+(a+b+c)=1  (using eq. (1))