A triangular plane ABC with centroid `(1,2,3)` cuts the coordinates axes at A, B, C, respectively
What is the equation of the plane ABC?

Options

A. `x+2y+3z=1`
B. `3x+2y+z=3`
C. `2x+3y+6z=18`
D. `6x+3y+2z=18`

Answer

To find the equation of the triangular plane ABC with a centroid at (1, 2, 3) that cuts the coordinate axes at points A, B, and C, we can follow these steps:

### Step 1: Define the points A, B, and C
Since the triangle cuts the coordinate axes, we can denote the points as follows:
- 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: \( C(0, 0, c) \)

### Step 2: Use the centroid formula
The centroid (G) of a triangle with vertices at \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), and \( C(x_3, y_3, z_3) \) is given by:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right) \]
Substituting the coordinates of points A, B, and C:
\[ G = \left( \frac{a + 0 + 0}{3}, \frac{0 + b + 0}{3}, \frac{0 + 0 + c}{3} \right) = \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right) \]

### Step 3: Set the centroid equal to the given coordinates
We know the centroid is given as \( (1, 2, 3) \). Therefore, we can set up the following equations:
\[ \frac{a}{3} = 1 \quad \Rightarrow \quad a = 3 \]
\[ \frac{b}{3} = 2 \quad \Rightarrow \quad b = 6 \]
\[ \frac{c}{3} = 3 \quad \Rightarrow \quad c = 9 \]

### Step 4: Write the intercept form of the plane equation
The equation of a plane in intercept form is given by:
\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]
Substituting the values of \( a, b, \) and \( c \):
\[ \frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1 \]

### Step 5: Clear the denominators
To eliminate the fractions, we can multiply through by the least common multiple of the denominators (which is 18):
\[ 18 \left( \frac{x}{3} + \frac{y}{6} + \frac{z}{9} \right) = 18 \]
This simplifies to:
\[ 6x + 3y + 2z = 18 \]

### Final Equation
Thus, the equation of the plane ABC is:
\[ 6x + 3y + 2z = 18 \]