To find the orthocenter of the triangle formed by the intersections of the plane \(2x + 3y + 6z = 12\) with the x-axis, y-axis, and z-axis, we will follow these steps:
### Step 1: Find the coordinates of points A, B, and C
1. **Point A (Intersection with x-axis)**: Set \(y = 0\) and \(z = 0\) in the plane equation:
\[
2x + 3(0) + 6(0) = 12 \implies 2x = 12 \implies x = 6
\]
Thus, \(A(6, 0, 0)\).
2. **Point B (Intersection with y-axis)**: Set \(x = 0\) and \(z = 0\):
\[
2(0) + 3y + 6(0) = 12 \implies 3y = 12 \implies y = 4
\]
Thus, \(B(0, 4, 0)\).
3. **Point C (Intersection with z-axis)**: Set \(x = 0\) and \(y = 0\):
\[
2(0) + 3(0) + 6z = 12 \implies 6z = 12 \implies z = 2
\]
Thus, \(C(0, 0, 2)\).
### Step 2: Determine the orthocenter of triangle ABC
The orthocenter of a triangle is the intersection of the altitudes. We will find the equations of the altitudes from each vertex.
1. **Altitude from A to line BC**:
- The coordinates of B and C are \(B(0, 4, 0)\) and \(C(0, 0, 2)\).
- The direction vector of line BC is \(\vec{BC} = C - B = (0 - 0, 0 - 4, 2 - 0) = (0, -4, 2)\).
- The altitude from A will be perpendicular to BC, so its direction vector will be \((0, -4, 2)\) (the normal to BC).
- The equation of the line through A(6, 0, 0) in the direction of \((0, -4, 2)\) is:
\[
(x, y, z) = (6, 0, 0) + t(0, -4, 2)
\]
Thus, the parametric equations are:
\[
x = 6, \quad y = -4t, \quad z = 2t
\]
2. **Altitude from B to line AC**:
- The coordinates of A and C are \(A(6, 0, 0)\) and \(C(0, 0, 2)\).
- The direction vector of line AC is \(\vec{AC} = C - A = (0 - 6, 0 - 0, 2 - 0) = (-6, 0, 2)\).
- The altitude from B will be perpendicular to AC, so its direction vector will be \((-6, 0, 2)\).
- The equation of the line through B(0, 4, 0) in the direction of \((-6, 0, 2)\) is:
\[
(x, y, z) = (0, 4, 0) + s(-6, 0, 2)
\]
Thus, the parametric equations are:
\[
x = -6s, \quad y = 4, \quad z = 2s
\]
3. **Altitude from C to line AB**:
- The coordinates of A and B are \(A(6, 0, 0)\) and \(B(0, 4, 0)\).
- The direction vector of line AB is \(\vec{AB} = B - A = (0 - 6, 4 - 0, 0 - 0) = (-6, 4, 0)\).
- The altitude from C will be perpendicular to AB, so its direction vector will be \((-6, 4, 0)\).
- The equation of the line through C(0, 0, 2) in the direction of \((-6, 4, 0)\) is:
\[
(x, y, z) = (0, 0, 2) + r(-6, 4, 0)
\]
Thus, the parametric equations are:
\[
x = -6r, \quad y = 4r, \quad z = 2
\]
### Step 3: Solve the equations to find the orthocenter
To find the orthocenter, we need to solve the equations of the altitudes simultaneously.
1. From the altitude from A:
\[
x = 6, \quad y = -4t, \quad z = 2t
\]
2. From the altitude from B:
\[
x = -6s, \quad y = 4, \quad z = 2s
\]
3. From the altitude from C:
\[
x = -6r, \quad y = 4r, \quad z = 2
\]
Setting \(x\) values equal:
\[
6 = -6s \implies s = -1
\]
Setting \(y\) values equal:
\[
-4t = 4 \implies t = -1
\]
Setting \(z\) values equal:
\[
2t = 2s \implies 2(-1) = 2(-1) \text{ (holds true)}
\]
Now substituting \(s = -1\) into the altitude from B:
\[
x = -6(-1) = 6, \quad y = 4, \quad z = 2(-1) = -2
\]
Thus, the orthocenter \(H\) is at:
\[
H(6, 4, -2)
\]
### Final Answer
The orthocenter of triangle ABC is \(H(6, 4, -2)\).
