Find the distance of the z-axis from the image of the point `M(2-3,3)` in the plane `x-2y-z+1=0.`

Explanation

To find the distance of the z-axis from the image of the point \( M(2, -3, 3) \) in the plane \( x - 2y - z + 1 = 0 \), we will follow these steps:

### Step 1: Identify the point and the plane equation
We have the point \( M(2, -3, 3) \) and the plane equation is given as:
\[ x - 2y - z + 1 = 0 \]

### Step 2: Find the image of the point \( M \) in the plane
To find the image of the point \( M \) in the plane, we will use the formula for the image of a point in a plane. The coordinates of the image \( M' (A_1, A_2, A_3) \) can be calculated using the following equations derived from the plane equation:

\[ \frac{A_1 - 2}{1} = \frac{A_2 + 3}{-2} = \frac{A_3 - 3}{-1} = -2d \]

Where \( d \) is the distance from the point to the plane.

### Step 3: Calculate the distance \( d \)
First, we need to calculate \( d \) using the formula for the distance from a point to a plane:
\[ d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]
For our plane \( x - 2y - z + 1 = 0 \), we have \( a = 1, b = -2, c = -1, d = 1 \) and the point \( M(2, -3, 3) \).

Calculating \( d \):
\[ d = \frac{|1 \cdot 2 + (-2) \cdot (-3) + (-1) \cdot 3 + 1|}{\sqrt{1^2 + (-2)^2 + (-1)^2}} = \frac{|2 + 6 - 3 + 1|}{\sqrt{1 + 4 + 1}} = \frac{|6|}{\sqrt{6}} = \frac{6}{\sqrt{6}} = \sqrt{6} \]

### Step 4: Substitute \( d \) back to find \( A_1, A_2, A_3 \)
Now substituting \( d \) back into the equations:
\[ A_1 - 2 = -2(-2\sqrt{6}) \implies A_1 - 2 = 4\sqrt{6} \implies A_1 = 4\sqrt{6} + 2 \]
\[ A_2 + 3 = -2(-2\sqrt{6}) \implies A_2 + 3 = 4\sqrt{6} \implies A_2 = 4\sqrt{6} - 3 \]
\[ A_3 - 3 = -2(-2\sqrt{6}) \implies A_3 - 3 = 4\sqrt{6} \implies A_3 = 4\sqrt{6} + 3 \]

### Step 5: Find the coordinates of the image point \( M' \)
Thus, the coordinates of the image point \( M' \) are:
\[ M' = (4\sqrt{6} + 2, 4\sqrt{6} - 3, 4\sqrt{6} + 3) \]

### Step 6: Calculate the distance from the z-axis
The distance from the z-axis is simply the x-coordinate of the image point \( M' \), which is \( 4\sqrt{6} + 2 \).

### Final Answer
The distance of the z-axis from the image of the point \( M(2, -3, 3) \) in the plane \( x - 2y - z + 1 = 0 \) is:
\[ \text{Distance} = 4\sqrt{6} + 2 \]