Match Conditions for System of Homogeneous Linear Equations

To find the geometric representation of the given system of linear equations:
$ax + by + cz = 0$
$bx + cy + az = 0$
$cx + ay + bz = 0$

Let $A$ be the coefficient matrix of the system:

$$A = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix}$$

The determinant of matrix $A$ is given by:

$$\Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = -(a^3 + b^3 + c^3 - 3abc) = -(a+b+c)(a^2+b^2+c^2 - ab - bc - ca)$$

Using the algebraic identity:
$a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}\left[(a-b)^2 + (b-c)^2 + (c-a)^2\right]$

Now we examine each of the conditions given in Column I:

1. Condition (A): $a + b + c \neq 0$ and $a^2 + b^2 + c^2 = ab + bc + ca$
   Since $a, b, c \in \mathbb{R}$,
   $\frac{1}{2}\left[(a-b)^2 + (b-c)^2 + (c-a)^2\right] = 0 \implies a = b = c$
   Since $a + b + c \neq 0$, we have $a = b = c \neq 0$.
   Substituting $a = b = c$ into the system of equations, each equation reduces to:
   $a(x + y + z) = 0 \implies x + y + z = 0$
   Thus, all three equations represent the same plane, i.e., identical planes.
   Therefore, (A) $\to$ (r).

2. Condition (B): $a + b + c = 0$ and $a^2 + b^2 + c^2 \neq ab + bc + ca$
   Here, $\Delta = 0$. Since $a^2 + b^2 + c^2 \neq ab + bc + ca$, $a, b,$ and $c$ are not all equal, meaning the normal vectors to the planes are not parallel and the planes are distinct.
   Adding the coordinates for any point on the line $x = y = z$ (i.e., $(k, k, k)$):
   $a(k) + b(k) + c(k) = k(a + b + c) = k(0) = 0$
   Similarly, the point satisfies the second and third equations.
   Since the rank of matrix $A$ is $2$, the solution space has dimension $3 - 2 = 1$, which is a line.
   Thus, the three planes intersect along the line $x = y = z$.
   Therefore, (B) $\to$ (q).

3. Condition (C): $a + b + c \neq 0$ and $a^2 + b^2 + c^2 \neq ab + bc + ca$
   In this case, neither factor of $\Delta$ is zero, which implies:
   $\Delta \neq 0$
   For a homogeneous system of linear equations with a non-zero determinant, there exists only the trivial solution:
   $(x, y, z) = (0, 0, 0)$
   Geometrically, the planes meet only at the origin (a single point).
   Therefore, (C) $\to$ (p).

4. Condition (D): $a + b + c = 0$ and $a^2 + b^2 + c^2 = ab + bc + ca$
   From $a^2 + b^2 + c^2 = ab + bc + ca$, we obtain $a = b = c$.
   Substituting this into $a + b + c = 0$ yields $3a = 0 \implies a = b = c = 0$.
   With $a = b = c = 0$, all three equations become:
   $0 \cdot x + 0 \cdot y + 0 \cdot z = 0$
   This equation is identically satisfied for all points $(x, y, z) \in \mathbb{R}^3$.
   Thus, the equations represent the whole of the three-dimensional space.
   Therefore, (D) $\to$ (s).

Final Matching:

• (A) $\to$ (r)
• (B) $\to$ (q)
• (C) $\to$ (p)
• (D) $\to$ (s)