Find the distance of the point (0.-3. -2) from the plane `x + 2y - z = 1` measured parallel to `(x+1)/2 = (y+1)/2 = z/3`

Answer

To find the distance of the point \( P(0, -3, -2) \) from the plane \( x + 2y - z = 1 \) measured parallel to the line given by the equations \( \frac{x + 1}{2} = \frac{y + 1}{2} = \frac{z}{3} \), we will follow these steps:

### Step 1: Identify the direction ratios of the line
The line is given in symmetric form. From the equations \( \frac{x + 1}{2} = \frac{y + 1}{2} = \frac{z}{3} \), we can identify the direction ratios \( l, m, n \) as:
- \( l = 2 \)
- \( m = 2 \)
- \( n = 3 \)

### Step 2: Write the parametric equations of the line through point \( P \)
The point \( P(0, -3, -2) \) will be used to write the parametric equations of the line parallel to the given line:
- \( x = 0 + 2\lambda = 2\lambda \)
- \( y = -3 + 2\lambda \)
- \( z = -2 + 3\lambda \)

### Step 3: Find a point on the plane
We need to find a point on the plane \( x + 2y - z = 1 \) that lies on the line we just defined. Substitute the parametric equations into the plane equation:
\[ 2\lambda + 2(-3 + 2\lambda) - (-2 + 3\lambda) = 1 \]
Simplifying this:
\[ 2\lambda - 6 + 4\lambda + 2 - 3\lambda = 1 \]
\[ (2\lambda + 4\lambda - 3\lambda) - 6 + 2 = 1 \]
\[ 3\lambda - 4 = 1 \]
\[ 3\lambda = 5 \implies \lambda = \frac{5}{3} \]

### Step 4: Calculate the coordinates of the point on the plane
Substituting \( \lambda = \frac{5}{3} \) back into the parametric equations:
- \( x = 2 \times \frac{5}{3} = \frac{10}{3} \)
- \( y = -3 + 2 \times \frac{5}{3} = -3 + \frac{10}{3} = -\frac{9}{3} + \frac{10}{3} = \frac{1}{3} \)
- \( z = -2 + 3 \times \frac{5}{3} = -2 + 5 = 3 \)

So the coordinates of the point \( N \) on the plane are \( N\left(\frac{10}{3}, \frac{1}{3}, 3\right) \).

### Step 5: Calculate the distance between points \( P \) and \( N \)
Now we will use the distance formula to find the distance \( D \) between points \( P(0, -3, -2) \) and \( N\left(\frac{10}{3}, \frac{1}{3}, 3\right) \):
\[ D = \sqrt{\left(\frac{10}{3} - 0\right)^2 + \left(\frac{1}{3} - (-3)\right)^2 + \left(3 - (-2)\right)^2} \]
Calculating each term:
- \( \left(\frac{10}{3}\right)^2 = \frac{100}{9} \)
- \( \left(\frac{1}{3} + 3\right)^2 = \left(\frac{1}{3} + \frac{9}{3}\right)^2 = \left(\frac{10}{3}\right)^2 = \frac{100}{9} \)
- \( (3 + 2)^2 = 5^2 = 25 \)

Now substituting back into the distance formula:
\[ D = \sqrt{\frac{100}{9} + \frac{100}{9} + 25} = \sqrt{\frac{100}{9} + \frac{100}{9} + \frac{225}{9}} = \sqrt{\frac{425}{9}} = \frac{\sqrt{425}}{3} \]
Since \( 425 = 25 \times 17 \), we can simplify:
\[ D = \frac{5\sqrt{17}}{3} \]

Thus, the distance of the point \( (0, -3, -2) \) from the plane \( x + 2y - z = 1 \) measured parallel to the given line is \( \frac{5\sqrt{17}}{3} \).