a, b, c are distinct real numbers, not equal to one. If ax + y + z = 0, x + by + z = 0, and x + y + cz = 0 have a non-trivial solution, then the value of 11−a+11−b+11−c is equal to

Since the system has non-trivial solution, we havea111b111c=0Applying R1→R1−R2,R2→R2−R3, we getΔ=a−11−b00b−11−c11c=0or c(1 - a)(1 - b) + (1 - b)(1 - c) - (1 - c)(a - 1) = 0Dividing throughout by (1 - a) (1 - b) (1 - c), we getc1−c+11−a+11−b=0or −1+11−c+11−a+11−b=0or 11−c+11−a+11−b=1