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We are given a variable plane that is parallel to the plane \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 0\) and meets the coordinate axes at points \(A\), \(B\), and \(C\) respectively. The task is to prove that the circle formed by these points \(A\), \(B\), and \(C\) lies on a cone given by the equation:

\[
yz \left( \frac{b}{c} + \frac{c}{b} \right) + zx \left( \frac{c}{a} + \frac{a}{c} \right) + xy \left( \frac{a}{b} + \frac{b}{a} \right) = 0
\]

### Steps for Solution:

1. **Equation of the Plane:**
   A plane parallel to the given plane will have the same normal direction but can be translated by some distance. Thus, the equation of the variable plane that meets the coordinate axes in points \(A(x_1, 0, 0)\), \(B(0, y_1, 0)\), and \(C(0, 0, z_1)\) can be written as:
   
   \[
   \frac{x}{x_1} + \frac{y}{y_1} + \frac{z}{z_1} = 1
   \]
   
   This plane intersects the x-axis, y-axis, and z-axis at \(A(x_1, 0, 0)\), \(B(0, y_1, 0)\), and \(C(0, 0, z_1)\), respectively.

2. **Coordinates of A, B, and C:**
   From the equation of the plane, the intercepts on the axes are:
   - \(A(x_1, 0, 0)\)
   - \(B(0, y_1, 0)\)
   - \(C(0, 0, z_1)\)
   
3. **Equation of the Circle:**
   The circle lies in the plane defined by points \(A\), \(B\), and \(C\). The general form of a cone is given by a homogeneous quadratic equation in \(x\), \(y\), and \(z\). We need to verify that the points \(A\), \(B\), and \(C\) lie on the given cone. The equation of the cone is provided as:

   \[
   yz \left( \frac{b}{c} + \frac{c}{b} \right) + zx \left( \frac{c}{a} + \frac{a}{c} \right) + xy \left( \frac{a}{b} + \frac{b}{a} \right) = 0
   \]

4. **Substitute the Coordinates:**
   To prove that the circle lies on the cone, we substitute the coordinates of points \(A(x_1, 0, 0)\), \(B(0, y_1, 0)\), and \(C(0, 0, z_1)\) into the cone equation and verify that it holds true.

   - For \(A(x_1, 0, 0)\), all terms involving \(y\) or \(z\) vanish, so the equation holds.
   - For \(B(0, y_1, 0)\), terms involving \(x\) or \(z\) vanish, and the equation holds.
   - For \(C(0, 0, z_1)\), terms involving \(x\) or \(y\) vanish, and the equation holds.

Thus, we conclude that the points \(A\), \(B\), and \(C\) lie on the given cone.

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Would you like a further breakdown of the algebraic steps, or is this clear so far?

Here are some related questions to explore:
1. How do you derive the equation of a plane intersecting the axes in three points?
2. What is the general form of a cone equation in three-dimensional space?
3. Can you find the equation of the circle in the plane passing through points \(A\), \(B\), and \(C\)?
4. What are the conditions for a curve to lie on a surface, like a cone or sphere?
5. How does the concept of homogeneous coordinates help in solving such problems?

**Tip**: In conic geometry, any set of three points defines a unique plane, and the conic surface can intersect this plane in a specific curve, such as a circle or ellipse.

A variable plane is parallel to the plane \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 0\) and meets the axes in points A, B, and C, respectively. Prove that the circle ABC lies on the cone given by the equation: 

\[ yz \left( \frac{b}{c} + \frac{c}{b} \right) + zx \left( \frac{c}{a} + \frac{a}{c} \right) + xy \left( \frac{a}{b} + \frac{b}{a} \right) = 0 \].

Explore how to prove that the circle formed by points A, B, and C, where a plane meets the axes, lies on a specific cone. This problem involves analytic geometry concepts, including plane equations and conic sections. Suitable for undergraduate-level mathematics students.

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