Find the perpendicular distance of the point `(1, 1, 1)` from the line `(x-2)/(2)=(y+3)/(2)=(z)/(-1)`.

Answer

To find the perpendicular distance of the point \( A(1, 1, 1) \) from the line given by the equations \( \frac{x-2}{2} = \frac{y+3}{2} = \frac{z}{-1} \), we will follow these steps:

### Step 1: Parametrize the line
The line can be expressed in parametric form. Let \( t \) be the parameter:
- \( x = 2t + 2 \)
- \( y = 2t - 3 \)
- \( z = -t \)

### Step 2: Define the point on the line
Let point \( B \) on the line be represented as \( B(2t + 2, 2t - 3, -t) \).

### Step 3: Find the direction ratios of line AB
The direction ratios of the line segment \( AB \) (from point \( A(1, 1, 1) \) to point \( B(2t + 2, 2t - 3, -t) \)) are given by:
- \( AB_x = (2t + 2) - 1 = 2t + 1 \)
- \( AB_y = (2t - 3) - 1 = 2t - 4 \)
- \( AB_z = -t - 1 = -t - 1 \)

### Step 4: Use the condition for perpendicularity
Since \( AB \) is perpendicular to the direction ratios of the line, we can use the direction ratios of the line, which are \( (2, 2, -1) \). The dot product of the direction ratios of \( AB \) and the line must equal zero:
\[ (2t + 1) \cdot 2 + (2t - 4) \cdot 2 + (-t - 1) \cdot (-1) = 0 \]

### Step 5: Simplify the equation
Expanding the equation:
\[ 4t + 2 + 4t - 8 + t + 1 = 0 \]
Combine like terms:
\[ 9t - 5 = 0 \]
Thus,
\[ t = \frac{5}{9} \]

### Step 6: Find the coordinates of point B
Substituting \( t = \frac{5}{9} \) back into the parametric equations of the line:
- \( x = 2 \cdot \frac{5}{9} + 2 = \frac{10}{9} + 2 = \frac{28}{9} \)
- \( y = 2 \cdot \frac{5}{9} - 3 = \frac{10}{9} - 3 = \frac{10}{9} - \frac{27}{9} = -\frac{17}{9} \)
- \( z = -\frac{5}{9} \)

Thus, the coordinates of point \( B \) are \( B\left(\frac{28}{9}, -\frac{17}{9}, -\frac{5}{9}\right) \).

### Step 7: Calculate the distance AB
Using the distance formula between points \( A(1, 1, 1) \) and \( B\left(\frac{28}{9}, -\frac{17}{9}, -\frac{5}{9}\right) \):
\[ AB = \sqrt{\left(\frac{28}{9} - 1\right)^2 + \left(-\frac{17}{9} - 1\right)^2 + \left(-\frac{5}{9} - 1\right)^2} \]
Calculating each term:
- \( \frac{28}{9} - 1 = \frac{28}{9} - \frac{9}{9} = \frac{19}{9} \)
- \( -\frac{17}{9} - 1 = -\frac{17}{9} - \frac{9}{9} = -\frac{26}{9} \)
- \( -\frac{5}{9} - 1 = -\frac{5}{9} - \frac{9}{9} = -\frac{14}{9} \)

Thus, the distance becomes:
\[ AB = \sqrt{\left(\frac{19}{9}\right)^2 + \left(-\frac{26}{9}\right)^2 + \left(-\frac{14}{9}\right)^2} \]
Calculating the squares:
\[ AB = \sqrt{\frac{361}{81} + \frac{676}{81} + \frac{196}{81}} = \sqrt{\frac{1233}{81}} = \frac{\sqrt{1233}}{9} \]

### Step 8: Simplify the distance
We can factor \( 1233 \) as \( 9 \times 137 \):
\[ AB = \frac{\sqrt{9 \times 137}}{9} = \frac{3\sqrt{137}}{9} = \frac{\sqrt{137}}{3} \]

### Final Answer
The perpendicular distance from the point \( (1, 1, 1) \) to the line is \( \frac{\sqrt{137}}{3} \) units.