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 \( P(1, 1, 1) \) from the line given by the equation

\[ \frac{x - 2}{2} = \frac{y + 3}{2} = \frac{z}{-1}, \]

we can follow these steps:

### Step 1: Parametrize the Line
The given line can be expressed in parametric form. Let \( t \) be the parameter. Then, we can write:

\[ x = 2t + 2, \]
\[ y = 2t - 3, \]
\[ z = -t. \]

### Step 2: Identify a Point on the Line
From the parametric equations, we can see that when \( t = 0 \), we have a point on the line:

\[ Q(2, -3, 0). \]

### Step 3: Find the Direction Ratios of the Line
The direction ratios of the line can be derived from the coefficients of \( t \) in the parametric equations:

\[ \text{Direction ratios} = (2, 2, -1). \]

### Step 4: Find the Vector \( \overrightarrow{PQ} \)
Next, we need to find the vector from point \( P(1, 1, 1) \) to point \( Q(2, -3, 0) \):

\[ \overrightarrow{PQ} = Q - P = (2 - 1, -3 - 1, 0 - 1) = (1, -4, -1). \]

### Step 5: Find the Projection of \( \overrightarrow{PQ} \) onto the Direction of the Line
To find the perpendicular distance, we need to project \( \overrightarrow{PQ} \) onto the direction vector of the line. The direction vector \( \mathbf{d} \) is \( (2, 2, -1) \).

The projection of \( \overrightarrow{PQ} \) onto \( \mathbf{d} \) is given by:

\[ \text{proj}_{\mathbf{d}} \overrightarrow{PQ} = \frac{\overrightarrow{PQ} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d}. \]

Calculating \( \overrightarrow{PQ} \cdot \mathbf{d} \):

\[ \overrightarrow{PQ} \cdot \mathbf{d} = (1)(2) + (-4)(2) + (-1)(-1) = 2 - 8 + 1 = -5. \]

Calculating \( \mathbf{d} \cdot \mathbf{d} \):

\[ \mathbf{d} \cdot \mathbf{d} = (2)(2) + (2)(2) + (-1)(-1) = 4 + 4 + 1 = 9. \]

Thus, the projection is:

\[ \text{proj}_{\mathbf{d}} \overrightarrow{PQ} = \frac{-5}{9} \mathbf{d} = \frac{-5}{9}(2, 2, -1) = \left(-\frac{10}{9}, -\frac{10}{9}, \frac{5}{9}\right). \]

### Step 6: Find the Vector Perpendicular to the Line
The vector perpendicular to the line is given by:

\[ \overrightarrow{PQ}_{\perp} = \overrightarrow{PQ} - \text{proj}_{\mathbf{d}} \overrightarrow{PQ}. \]

Calculating this:

\[ \overrightarrow{PQ}_{\perp} = (1, -4, -1) - \left(-\frac{10}{9}, -\frac{10}{9}, \frac{5}{9}\right) = \left(1 + \frac{10}{9}, -4 + \frac{10}{9}, -1 - \frac{5}{9}\right). \]

Converting to a common denominator:

\[ = \left(\frac{9}{9} + \frac{10}{9}, -\frac{36}{9} + \frac{10}{9}, -\frac{9}{9} - \frac{5}{9}\right) = \left(\frac{19}{9}, -\frac{26}{9}, -\frac{14}{9}\right). \]

### Step 7: Calculate the Magnitude of the Perpendicular Vector
The magnitude of this vector gives the perpendicular distance:

\[ d = \sqrt{\left(\frac{19}{9}\right)^2 + \left(-\frac{26}{9}\right)^2 + \left(-\frac{14}{9}\right)^2}. \]

Calculating each term:

\[ = \sqrt{\frac{361}{81} + \frac{676}{81} + \frac{196}{81}} = \sqrt{\frac{1233}{81}} = \frac{1}{9} \sqrt{1233}. \]

### Final Answer
Thus, the perpendicular distance from the point \( P(1, 1, 1) \) to the line is:

\[ \frac{1}{9} \sqrt{1233}. \]