Key Concepts and Formulas

This problem draws upon fundamental concepts from 3D Geometry:

1. Equation of a Plane in Intercept Form: If a plane intercepts the x-axis at $(a,0,0)$, the y-axis at $(0,b,0)$, and the z-axis at $(0,0,c)$, its equation is given by: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ Here, $a, b, c$ are the x, y, and z intercepts, respectively. This form is particularly useful when the points of intersection with the axes are known.

2. Centroid of a Triangle in 3D: For a triangle with vertices $P_1(x_1, y_1, z_1)$, $P_2(x_2, y_2, z_2)$, and $P_3(x_3, y_3, z_3)$, the coordinates of its centroid $G$ are the average of the respective coordinates: $G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$

3. Distance of a Plane from the Origin: The perpendicular distance $d$ from the origin $(0,0,0)$ to a plane represented by the general equation $Ax + By + Cz + D = 0$ is given by the formula: $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$

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Step-by-Step Solution

1. Define the Plane and its Intercepts

Let the variable plane intersect the coordinate axes at points A, B, and C. Since these points lie on the respective axes, we can define their coordinates in terms of the plane's intercepts. Let the intercepts be $a, b, c$ on the x, y, and z axes, respectively.

• Point A on the x-axis: $A = (a, 0, 0)$
• Point B on the y-axis: $B = (0, b, 0)$
• Point C on the z-axis: $C = (0, 0, c)$

Using the intercept form, the equation of this variable plane is: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \quad \text{(Equation 1)}$ Explanation: We choose the intercept form because the problem directly provides information about the plane's intersections with the coordinate axes, making this the most natural and efficient representation for the plane.

2. Express the Centroid of $\Delta ABC$ in terms of Intercepts

Let the coordinates of the centroid of $\Delta ABC$ be $G(h, k, l)$. Our goal is to find the locus of this point, so we will express $h, k, l$ in terms of the intercepts $a, b, c$. Using the centroid formula for the vertices $A(a,0,0)$, $B(0,b,0)$, $C(0,0,c)$: $h = \frac{a+0+0}{3} \implies a = 3h$ $k = \frac{0+b+0}{3} \implies b = 3k$ $l = \frac{0+0+c}{3} \implies c = 3l$ Explanation: This step establishes a direct relationship between the parameters of the plane (its intercepts $a,b,c$) and the coordinates of the centroid $(h,k,l)$. This is a critical link for eventually deriving the locus of the centroid.

3. Substitute Centroid Coordinates into the Plane Equation

Now, we substitute the expressions for $a, b, c$ (obtained in Step 2) back into Equation 1. This will give us the equation of the variable plane, but now expressed in terms of the centroid's coordinates $(h,k,l)$: $\frac{x}{3h} + \frac{y}{3k} + \frac{z}{3l} = 1 \quad \text{(Equation 2)}$ Explanation: By making this substitution, we have translated the problem from dealing with arbitrary intercepts to dealing with the coordinates of the centroid. This equation now describes the plane passing through points whose centroid is $(h,k,l)$.

4. Utilize the Given Distance from the Origin

The problem states that the plane is at a constant distance of 3 units from the origin $(0,0,0)$. To apply the distance formula, we first rewrite Equation 2 in the general form $Ax + By + Cz + D = 0$: $\left(\frac{1}{3h}\right)x + \left(\frac{1}{3k}\right)y + \left(\frac{1}{3l}\right)z - 1 = 0$ Here, we identify the coefficients as $A = \frac{1}{3h}$, $B = \frac{1}{3k}$, $C = \frac{1}{3l}$, and $D = -1$. The given distance $d$ is 3. Applying the distance formula from the origin: $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$ Substituting the values: $3 = \frac{|-1|}{\sqrt{\left(\frac{1}{3h}\right)^2 + \left(\frac{1}{3k}\right)^2 + \left(\frac{1}{3l}\right)^2}}$ $3 = \frac{1}{\sqrt{\frac{1}{9h^2} + \frac{1}{9k^2} + \frac{1}{9l^2}}}$ Explanation: This step integrates the final piece of information given in the problem: the constant distance of the plane from the origin. By applying the distance formula, we establish a direct algebraic relationship involving the centroid's coordinates $(h,k,l)$ and the constant distance.

5. Simplify the Equation to Find the Locus

To simplify the equation and find the locus, we will perform algebraic manipulations: First, square both sides of the equation from Step 4: $3^2 = \left(\frac{1}{\sqrt{\frac{1}{9h^2} + \frac{1}{9k^2} + \frac{1}{9l^2}}}\right)^2$ $9 = \frac{1}{\frac{1}{9h^2} + \frac{1}{9k^2} + \frac{1}{9l^2}}$ Next, take the reciprocal of both sides to bring the terms with $h,k,l$ to the numerator: $\frac{1}{9} = \frac{1}{9h^2} + \frac{1}{9k^2} + \frac{1}{9l^2}$ Finally, multiply the entire equation by 9 to clear the denominators: $9 \times \frac{1}{9} = 9 \left( \frac{1}{9h^2} + \frac{1}{9k^2} + \frac{1}{9l^2} \right)$ $1 = \frac{9}{9h^2} + \frac{9}{9k^2} + \frac{9}{9l^2}$ $1 = \frac{1}{h^2} + \frac{1}{k^2} + \frac{1}{l^2}$ To express the locus of the centroid, we replace the general coordinates $(h, k, l)$ with $(x, y, z)$: $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 1$ Explanation: The algebraic simplification isolates the relationship between the coordinates of the centroid. The final substitution of $(h,k,l)$ with $(x,y,z)$ is the standard convention for representing the equation of a locus.

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Tips and Common Mistakes to Avoid

• Intercept Form vs. Normal Form: Be careful not to confuse the intercept form of a plane's equation with the normal form. The intercept form is ideal when dealing with axis intersections.
• Centroid Definition: Remember that the centroid is simply the average of the coordinates of the vertices. For 3D, this applies to x, y, and z coordinates independently.
• Distance Formula: Double-check the distance formula for a plane from the origin. Ensure the absolute value of $D$ is taken in the numerator and the square root of the sum of squares of coefficients $A, B, C$ is in the denominator.
• Algebraic Errors: Squaring both sides and taking reciprocals are common steps in locus problems. Be meticulous with these algebraic manipulations to avoid sign errors or incorrect distributions.
• Locus Convention: Always substitute the general variables $(x,y,z)$ for the specific coordinates of the variable point (like $h,k,l$) in the final step to correctly represent the locus.

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Summary and Key Takeaway

This problem effectively combines several core concepts from 3D geometry to determine the locus of a point. The strategy involved:

1. Representing the variable plane using its intercepts.
2. Connecting these intercepts to the coordinates of the centroid of the triangle formed by the intercepts.
3. Utilizing the given constraint of the plane's distance from the origin to form an equation solely in terms of the centroid's coordinates.
4. Simplifying this equation algebraically to arrive at the final locus.

The resulting locus $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 1$ describes the surface on which the centroid of $\Delta ABC$ must lie, given the condition that the plane is always 3 units away from the origin.

The final answer is $\boxed{\text{(A)}}$.