Key Concepts and Formulas

1. Equation of a Plane in Intercept Form: If a plane intersects the x, y, and z axes at points A($a$, 0, 0), B(0, $b$, 0), and C(0, 0, $c$) respectively, its equation can be written as: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ Here, $a, b, c$ are the x, y, and z-intercepts.

2. Centroid of a Triangle in 3D: For a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$, its centroid G is calculated as the average of the 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. Equation of a Line Perpendicular to a Plane: The normal vector to a plane $Ax + By + Cz + D = 0$ is $\vec{n} = (A, B, C)$. A line perpendicular to this plane will have its direction vector parallel to the plane's normal vector. If the line passes through a point $(x_0, y_0, z_0)$ and has a direction vector $(l, m, n)$, its symmetric equation is: $\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}$

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Step-by-Step Solution

Step 1: Define the Coordinates of Vertices A, B, C and the General Equation of Plane P

• Why this step? The problem states that plane P meets the coordinate axes at A, B, and C. To utilize the centroid information, we first need to express the coordinates of these intersection points and the plane's equation in terms of its intercepts.
• Let the plane P make intercepts $a, b, c$ with the x, y, and z axes, respectively.
• Based on the intercept definition, the coordinates of the points are:
  • A = ($a$, 0, 0) (on the x-axis)
  • B = (0, $b$, 0) (on the y-axis)
  • C = (0, 0, $c$) (on the z-axis)
• The general equation of plane P in intercept form is: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

Step 2: Use the Centroid Information to Determine the Intercepts $a, b, c$

• Why this step? We are given the centroid of $\Delta ABC$ and have expressed the vertices A, B, C in terms of $a, b, c$. By applying the centroid formula and equating it to the given centroid, we can solve for the specific values of $a, b, c$.
• Using the centroid formula for $\Delta ABC$ with vertices A($a$, 0, 0), B(0, $b$, 0), C(0, 0, $c$): $G = \left( \frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3} \right) = \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right)$
• The problem states that the centroid of $\Delta ABC$ is (1, 1, 2).
• Equating the coordinates of the calculated centroid with the given centroid: $\frac{a}{3} = 1 \quad \Rightarrow \quad a = 3$ $\frac{b}{3} = 1 \quad \Rightarrow \quad b = 3$ $\frac{c}{3} = 2 \quad \Rightarrow \quad c = 6$

Step 3: Find the Specific Equation of Plane P and its Normal Vector

• Why this step? Having found the intercepts $a, b, c$, we can now write the precise equation of plane P. From this equation, we can directly identify the normal vector, which is crucial for determining the direction of a line perpendicular to the plane.
• Substitute the values $a=3, b=3, c=6$ into the intercept form of the plane equation: $\frac{x}{3} + \frac{y}{3} + \frac{z}{6} = 1$
• To convert this into the general form $Ax + By + Cz = D$, we can multiply the entire equation by the least common multiple of the denominators (which is 6): $6 \left( \frac{x}{3} \right) + 6 \left( \frac{y}{3} \right) + 6 \left( \frac{z}{6} \right) = 6 \times 1$ $2x + 2y + z = 6$
• From the general form $2x + 2y + z - 6 = 0$, the normal vector to plane P is $\vec{n} = (2, 2, 1)$.

Step 4: Determine the Equation of the Line Perpendicular to Plane P and Passing Through the Centroid

• Why this step? We need to find the equation of a line. For this, we require a point on the line and its direction vector. We are given the point (the centroid), and we have just found the direction vector (the normal to the plane).
• The line passes through the centroid G = (1, 1, 2). So, $(x_0, y_0, z_0) = (1, 1, 2)$.
• A line perpendicular to plane P will have its direction vector parallel to the normal vector of plane P.
• The normal vector of plane P is $\vec{n} = (2, 2, 1)$. Therefore, the direction vector of the required line is $(l, m, n) = (2, 2, 1)$.
• Using the symmetric form of the line equation: $\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}$ $\frac{x - 1}{2} = \frac{y - 1}{2} = \frac{z - 2}{1}$

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Common Mistakes & Tips

1. Converting Plane Forms: Always ensure you can correctly convert between the intercept form and the general form of a plane equation, as the normal vector is most easily identified from the general form.
2. Centroid Calculation: Double-check the centroid calculation, especially when some coordinates are zero. It's a simple average, but errors can occur.
3. Normal vs. Direction Vectors: Remember that a line perpendicular to a plane uses the plane's normal vector as its direction vector. If a question asks for a line parallel to a plane, its direction vector must be perpendicular to the plane's normal vector (their dot product would be zero).
4. Symmetric Form of Line: Be careful with the signs when plugging in the point coordinates $(x_0, y_0, z_0)$ into the symmetric form $\frac{x - x_0}{l}$.

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Summary

This problem elegantly combines several fundamental concepts of 3D geometry. We began by defining the plane and its axis intercepts using the intercept form. The given centroid of the triangle formed by these intercepts allowed us to precisely determine the values of the intercepts. With the specific plane equation, we extracted its normal vector. Finally, using the given centroid as a point on the line and the plane's normal vector as the line's direction vector, we constructed the symmetric equation of the required perpendicular line.

The final answer is $\boxed{\text{x − 1 2 = y − 1 2 = z − 2 1}}$, which corresponds to option (A).