The coordinates of the centriod of triangle ABC where A,B,C are the points of intersection of the plane `6x + 3y -2z =18` with the coordinate axes are :

Options

A. `(1,2,-3)`
B. `(-1, 2,3)`
C. `(-1,-2,-3)`
D. `(1,-2,3)`

Answer

To find the coordinates of the centroid of triangle ABC, where A, B, and C are the points of intersection of the plane \(6x + 3y - 2z = 18\) with the coordinate axes, we can follow these steps:

### Step 1: Find the intercepts of the plane with the coordinate axes.
To find the points of intersection, we can set two of the variables to zero and solve for the third.

1. **Finding point A (x-intercept)**:
Set \(y = 0\) and \(z = 0\) in the equation:
\[ 6x + 3(0) - 2(0) = 18 \implies 6x = 18 \implies x = 3 \]
Thus, point A is \((3, 0, 0)\).

2. **Finding point B (y-intercept)**:
Set \(x = 0\) and \(z = 0\):
\[ 6(0) + 3y - 2(0) = 18 \implies 3y = 18 \implies y = 6 \]
Thus, point B is \((0, 6, 0)\).

3. **Finding point C (z-intercept)**:
Set \(x = 0\) and \(y = 0\):
\[ 6(0) + 3(0) - 2z = 18 \implies -2z = 18 \implies z = -9 \]
Thus, point C is \((0, 0, -9)\).

### Step 2: Calculate the coordinates of the centroid.
The centroid \(G\) of a triangle with vertices at points \(A(x_1, y_1, z_1)\), \(B(x_2, y_2, z_2)\), and \(C(x_3, y_3, z_3)\) is given by 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:
- \(A(3, 0, 0)\)
- \(B(0, 6, 0)\)
- \(C(0, 0, -9)\)

We find the coordinates of the centroid:
\[ G\left(\frac{3 + 0 + 0}{3}, \frac{0 + 6 + 0}{3}, \frac{0 + 0 - 9}{3}\right) = G\left(\frac{3}{3}, \frac{6}{3}, \frac{-9}{3}\right) \]
Calculating each component:
\[ G(1, 2, -3) \]

### Final Result:
The coordinates of the centroid of triangle ABC are \((1, 2, -3)\).

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