A triangular plane ABC with centroid (1,2,3) cuts the coordinate axes at A, B, C respectivley.
What are the intercepts made by the plane ABC on the axes ?

Options

A. 3,6,9
B. 1,2,3
C. 1,4,9
D. 2,4,6

Explanation

To find the intercepts made by the triangular plane ABC on the coordinate axes, we will follow these steps:

### Step 1: Define the coordinates of points A, B, and C
Since the plane cuts the coordinate axes at points A, B, and C, we can denote:
- Point A (x-intercept) as \( A(x, 0, 0) \)
- Point B (y-intercept) as \( B(0, y, 0) \)
- Point C (z-intercept) as \( C(0, 0, z) \)

### Step 2: Use the centroid formula
The centroid (G) of a triangle formed by points A, B, and C is given by the formula:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right) \]
Substituting the coordinates of points A, B, and C into the formula, we have:
\[ G = \left( \frac{x + 0 + 0}{3}, \frac{0 + y + 0}{3}, \frac{0 + 0 + z}{3} \right) = \left( \frac{x}{3}, \frac{y}{3}, \frac{z}{3} \right) \]

### Step 3: Set the centroid equal to the given coordinates
We know from the problem statement that the centroid is at the point (1, 2, 3). Therefore, we can set up the following equations:
\[ \frac{x}{3} = 1 \quad (1) \]
\[ \frac{y}{3} = 2 \quad (2) \]
\[ \frac{z}{3} = 3 \quad (3) \]

### Step 4: Solve for x, y, and z
Now, we will solve each equation for x, y, and z.

From equation (1):
\[ x = 3 \times 1 = 3 \]

From equation (2):
\[ y = 3 \times 2 = 6 \]

From equation (3):
\[ z = 3 \times 3 = 9 \]

### Step 5: State the intercepts
Thus, the intercepts made by the plane ABC on the axes are:
- x-intercept (A) = 3
- y-intercept (B) = 6
- z-intercept (C) = 9

### Final Answer
The intercepts made by the plane ABC on the axes are:
- A (3, 0, 0)
- B (0, 6, 0)
- C (0, 0, 9)

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