To find the distance of the point \( P(1, -2, 3) \) from the plane \( x - y + z = 5 \) measured along a line parallel to the direction given by \( \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} \), we can follow these steps:
### Step 1: Identify the Direction Ratios
The line is given in the form \( \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} \). The direction ratios of this line can be taken as \( (2, 3, -6) \).
### Step 2: Write the Parametric Equations of the Line
Since the line passes through the point \( P(1, -2, 3) \), we can write the parametric equations of the line as:
\[
x = 1 + 2t
\]
\[
y = -2 + 3t
\]
\[
z = 3 - 6t
\]
where \( t \) is a parameter.
### Step 3: Find the Point on the Plane
To find the point \( Q \) on the plane \( x - y + z = 5 \), we substitute the parametric equations into the plane equation:
\[
(1 + 2t) - (-2 + 3t) + (3 - 6t) = 5
\]
Simplifying this, we get:
\[
1 + 2t + 2 - 3t + 3 - 6t = 5
\]
\[
6 - 7t = 5
\]
\[
-7t = -1 \implies t = \frac{1}{7}
\]
### Step 4: Calculate the Coordinates of Point \( Q \)
Now we can substitute \( t = \frac{1}{7} \) back into the parametric equations to find the coordinates of point \( Q \):
\[
x = 1 + 2\left(\frac{1}{7}\right) = 1 + \frac{2}{7} = \frac{9}{7}
\]
\[
y = -2 + 3\left(\frac{1}{7}\right) = -2 + \frac{3}{7} = -\frac{11}{7}
\]
\[
z = 3 - 6\left(\frac{1}{7}\right) = 3 - \frac{6}{7} = \frac{15}{7}
\]
Thus, the coordinates of point \( Q \) are \( \left(\frac{9}{7}, -\frac{11}{7}, \frac{15}{7}\right) \).
### Step 5: Calculate the Distance \( PQ \)
Now we can find the distance \( PQ \) using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]
Substituting the coordinates of 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 component:
\[
\frac{9}{7} - 1 = \frac{2}{7} \quad \Rightarrow \quad \left(\frac{2}{7}\right)^2 = \frac{4}{49}
\]
\[
-\frac{11}{7} + 2 = -\frac{11}{7} + \frac{14}{7} = \frac{3}{7} \quad \Rightarrow \quad \left(\frac{3}{7}\right)^2 = \frac{9}{49}
\]
\[
\frac{15}{7} - 3 = \frac{15}{7} - \frac{21}{7} = -\frac{6}{7} \quad \Rightarrow \quad \left(-\frac{6}{7}\right)^2 = \frac{36}{49}
\]
Now substituting back into the distance formula:
\[
d = \sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}} = \sqrt{\frac{49}{49}} = \sqrt{1} = 1
\]
### Final Result
The distance of the point \( (1, -2, 3) \) from the plane \( x - y + z = 5 \) measured along the line parallel to \( \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} \) is \( 1 \).
