Find the distance of the point (1,-2,3) from the plane x - y + z = 5, measured parallel to the line :
`(x)/(2) = (y)/(3) = (z)/(-6)` .

Answer

To find the distance of the point \( P(1, -2, 3) \) from the plane given by the equation \( x - y + z = 5 \), measured parallel to the line defined by the direction ratios \( \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} \), we can follow these steps:

### Step 1: Identify the direction ratios of the line
The line is given in symmetric form as:
\[ \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} \]
From this, we can identify the direction ratios of the line as \( \vec{d} = (2, 3, -6) \).

### Step 2: Parametrize the line through point \( P \)
Using the direction ratios, we can write the parametric equations of the line passing through point \( P(1, -2, 3) \):
\[ x = 2t + 1, \quad y = 3t - 2, \quad z = -6t + 3 \]
where \( t \) is a parameter.

### Step 3: Find the general point on the line
The general point \( Q \) on the line can be represented as:
\[ Q(t) = (2t + 1, 3t - 2, -6t + 3) \]

### Step 4: Substitute \( Q(t) \) into the plane equation
We need to find the value of \( t \) for which point \( Q(t) \) lies on the plane \( x - y + z = 5 \):
\[ (2t + 1) - (3t - 2) + (-6t + 3) = 5 \]
Simplifying this gives:
\[ 2t + 1 - 3t + 2 - 6t + 3 = 5 \]
\[ -7t + 6 = 5 \]
\[ -7t = -1 \quad \Rightarrow \quad t = \frac{1}{7} \]

### Step 5: Find the coordinates of point \( Q \)
Now substitute \( t = \frac{1}{7} \) back into the parametric equations to find the coordinates of point \( Q \):
\[ x = 2\left(\frac{1}{7}\right) + 1 = \frac{2}{7} + \frac{7}{7} = \frac{9}{7} \]
\[ y = 3\left(\frac{1}{7}\right) - 2 = \frac{3}{7} - \frac{14}{7} = -\frac{11}{7} \]
\[ z = -6\left(\frac{1}{7}\right) + 3 = -\frac{6}{7} + \frac{21}{7} = \frac{15}{7} \]
Thus, the coordinates of point \( Q \) are \( Q\left(\frac{9}{7}, -\frac{11}{7}, \frac{15}{7}\right) \).

### Step 6: Calculate the distance \( PQ \)
Now, we can calculate the distance \( d \) between points \( P(1, -2, 3) \) and \( Q\left(\frac{9}{7}, -\frac{11}{7}, \frac{15}{7}\right) \):
\[ d = \sqrt{\left(\frac{9}{7} - 1\right)^2 + \left(-\frac{11}{7} + 2\right)^2 + \left(\frac{15}{7} - 3\right)^2} \]
Calculating each term:
1. \( \frac{9}{7} - 1 = \frac{9}{7} - \frac{7}{7} = \frac{2}{7} \)
2. \( -\frac{11}{7} + 2 = -\frac{11}{7} + \frac{14}{7} = \frac{3}{7} \)
3. \( \frac{15}{7} - 3 = \frac{15}{7} - \frac{21}{7} = -\frac{6}{7} \)

Now substituting these values into the distance formula:
\[ d = \sqrt{\left(\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(-\frac{6}{7}\right)^2} \]
\[ = \sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}} = \sqrt{\frac{49}{49}} = \sqrt{1} = 1 \]

### Final Answer
The distance of the point \( (1, -2, 3) \) from the plane \( x - y + z = 5 \), measured parallel to the line, is \( 1 \).

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