IIT JEE Advanced 2007 Paper 1

We have
ax + by + cz = 0
bx + cy + az = 0
cx + ay + bz = 0
Δ = |

a	b	c
b	c	a
c	a	b

|
= - (a + b + c) (a2+b2+c2 - ab - bc - ac)
= −

1

2

(a + b + c) [(a−b)2+(b−c)2+(c−a)2]
(A)→(R)
a + b + c ≠ 0
a2+b2+c2 = ab + bc + ca
⇒ a = b = c ≠ 0
which implies that the equations represent identical planes.
(B)→(Q)
a + b + c = 0
and a2+b2+c2 ≠ ab + bc + ca
⇒ Δ = 0
which implies that the equations has infinitely many solution and
x = y = z = λ ∊ R
satisfy the given equation.
(C)→(P)
a + b + c ≠ 0
a2+b2+c2 ≠ ab + bc + ca
⇒ Δ ≠ 0
That is the equations has a unique solution.
(D)→(S)
a + b + c = 0
a2+b2+c2 = ab + bc + ac
⇒ a = b = c = 0
That is, all values of (x, y, z) satisfy the given equations and the equations represent whole of the three-dimensional space.