Does `(a)/(x-y)+(b)/(y-z)+(c)/(z-x)=0` represents a pair of planes?

Answer

To determine if the equation

\[ \frac{a}{x-y} + \frac{b}{y-z} + \frac{c}{z-x} = 0 \]

represents a pair of planes, we can follow these steps:

### Step 1: Rewrite the Equation
We start by rewriting the given equation in a more manageable form. We can multiply through by the denominators to eliminate the fractions:

\[ a(y-z)(z-x) + b(z-x)(x-y) + c(x-y)(y-z) = 0 \]

### Step 2: Expand the Terms
Next, we expand each term in the equation:

1. \( a(y-z)(z-x) = a(yz - yx - z^2 + zx) \)
2. \( b(z-x)(x-y) = b(zx - zy - x^2 + xy) \)
3. \( c(x-y)(y-z) = c(xy - xz - y^2 + yz) \)

Combining these, we have:

\[ ayz - ayx - az^2 + azx + bzx - bzy - bx^2 + bxy + cxy - cxz - cy^2 + cyz = 0 \]

### Step 3: Collect Like Terms
Now we collect like terms based on powers of \(x\), \(y\), and \(z\):

\[ (ax^2 + by^2 + cz^2) + (2a - b - c)xy + (2b - a - c)xz + (2c - a - b)yz = 0 \]

### Step 4: Formulate the General Equation of Pair of Planes
The general equation of a pair of planes can be expressed as:

\[ A x^2 + B y^2 + C z^2 + 2F yz + 2G zx + 2H xy = 0 \]

### Step 5: Compare Coefficients
From our collected terms, we can identify:

- \(A = 0\) (since there is no \(x^2\) term)
- \(B = 0\) (since there is no \(y^2\) term)
- \(C = 0\) (since there is no \(z^2\) term)
- \(F = \frac{(2a - b - c)}{2}\)
- \(G = \frac{(2b - a - c)}{2}\)
- \(H = \frac{(2c - a - b)}{2}\)

### Step 6: Check the Condition for Pair of Planes
For the equation to represent a pair of planes, the following condition must hold:

\[ ABC + 2FGH - AF^2 - BG^2 - CH^2 = 0 \]

Since \(A = 0\), \(B = 0\), and \(C = 0\), we substitute these values into the condition:

\[ 0 + 2FGH - 0 - 0 - 0 = 0 \implies 2FGH = 0 \]

This means at least one of \(F\), \(G\), or \(H\) must be zero.

### Conclusion
If \(F\), \(G\), or \(H\) is zero, then the original equation represents a pair of planes. Therefore, the answer to the question is:

**Yes, the equation represents a pair of planes if the conditions are satisfied.**

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