To solve the problem, we need to find the equation of a line that is perpendicular to the plane formed by the triangle ABC and passes through the centroid G(1, 1, 2).
### Step-by-Step Solution:
1. **Identify Points A, B, and C:**
- The plane intersects the x-axis at point A, the y-axis at point B, and the z-axis at point C.
- Therefore, we can denote the coordinates of these points as:
- A = (a, 0, 0)
- B = (0, b, 0)
- C = (0, 0, c)
2. **Centroid of Triangle ABC:**
- The centroid G of triangle ABC 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 A, B, and C into the formula, we have:
\[
G\left(\frac{a + 0 + 0}{3}, \frac{0 + b + 0}{3}, \frac{0 + 0 + c}{3}\right) = \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)
\]
3. **Set the Centroid Equal to Given Point:**
- We know that G(1, 1, 2). Therefore, we can set up the following equations:
\[
\frac{a}{3} = 1 \implies a = 3
\]
\[
\frac{b}{3} = 1 \implies b = 3
\]
\[
\frac{c}{3} = 2 \implies c = 6
\]
4. **Coordinates of Points A, B, and C:**
- Now we have the coordinates of the points:
- A = (3, 0, 0)
- B = (0, 3, 0)
- C = (0, 0, 6)
5. **Equation of the Plane:**
- The equation of the plane can be derived from the intercepts on the axes:
\[
\frac{x}{3} + \frac{y}{3} + \frac{z}{6} = 1
\]
- Multiplying through by 6 gives:
\[
2x + 2y + z = 6
\]
6. **Direction Ratios of the Normal to the Plane:**
- The coefficients of x, y, and z in the plane equation give the direction ratios of the normal to the plane:
- Direction ratios = (2, 2, 1)
7. **Equation of the Line Perpendicular to the Plane:**
- The line that is perpendicular to the plane and passes through point G(1, 1, 2) can be expressed in symmetric form:
\[
\frac{x - 1}{2} = \frac{y - 1}{2} = \frac{z - 2}{1}
\]
### Final Answer:
The equation of the line perpendicular to the plane and passing through G is:
\[
\frac{x - 1}{2} = \frac{y - 1}{2} = z - 2
\]
