Correct Option: C

Given set of equations are,
x + y + z = 1, a – ay + 32 = 5 and 5x – 3y + az = 6
Now ρ(A) = ρ(AB) < no. of variables
| A| = 0

	1			1			1		= 0
	a			-a			3
	5			-3			a

(– a² + 9) – (a² – 15) + (– 3a + 5a) = 0
9 – a² + 15 – a² + 2a = 0
2a² – 2a – 24 = 0
a² – a – 12 = 0
a² – 4a + 3a – 12 = 0
(a = 4, – 3)
ρ(AB) < (No. of variable)

	1			1			1		= 0
	-a			3			5
	-3			a			6

1(18 – 5a) – 1(– 6a + 15) + 1(– a² + 9) = 0
18 – 5a – 15 + 6a + 9 – a² = 0
a² – a – 12 = 0
a = 3, 4 or a = 4

	1			1			1		= 0
	a			-a			5
	5			-3			6

(– 6a + 15) – (6a – 25) + (– 3a + 5a) = 0
– 6a + 15 – 6a + 25 + 2a = 0
10a = 40
a = 4
if a = 4
| AB| = 0

	1			1			1		= 0
	a			3			5
	5			a			6

(18 – 5a) – (6a – 25) + (a² – 15) = 0
a² – 11a + 28 = 0
a = 7, 4 or a = 4