The point of interesection of the line x = y = z with the plane x + 2y + 3z = 6 is (1,1,-1).

Answer

To determine whether the point of intersection of the line \( x = y = z \) with the plane \( x + 2y + 3z = 6 \) is indeed \( (1, 1, -1) \), we can follow these steps:

### Step 1: Understand the line equation
The line is defined by the equations \( x = y = z \). We can express this in terms of a parameter \( \lambda \):
\[ x = \lambda, \quad y = \lambda, \quad z = \lambda \]
Thus, any point on the line can be represented as \( (\lambda, \lambda, \lambda) \).

### Step 2: Substitute into the plane equation
Next, we substitute \( x, y, z \) from the line into the plane equation \( x + 2y + 3z = 6 \):
\[ \lambda + 2\lambda + 3\lambda = 6 \]
This simplifies to:
\[ 6\lambda = 6 \]

### Step 3: Solve for \( \lambda \)
Now, we solve for \( \lambda \):
\[ \lambda = 1 \]

### Step 4: Find the coordinates of the intersection point
Substituting \( \lambda = 1 \) back into the equations for the line gives us the coordinates of the intersection point:
\[ x = 1, \quad y = 1, \quad z = 1 \]
Thus, the point of intersection is \( (1, 1, 1) \).

### Step 5: Compare with the given point
The given point of intersection is \( (1, 1, -1) \). Since our calculated point of intersection is \( (1, 1, 1) \), we can conclude that the statement is false.

### Conclusion
The point of intersection of the line \( x = y = z \) with the plane \( x + 2y + 3z = 6 \) is \( (1, 1, 1) \), not \( (1, 1, -1) \).

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