To find the distance of the point \( (1, -2, 3) \) from the plane \( x - y + z = 5 \) measured parallel to the line given by \( \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} \), we can follow these steps:
### Step 1: Identify the point and the plane equation
- The point \( P \) is given as \( (1, -2, 3) \).
- The equation of the plane is \( x - y + z = 5 \).
### Step 2: Find the direction ratios of the line
The line is given in symmetric form:
\[
\frac{x}{2} = \frac{y}{3} = \frac{z}{-6}
\]
From this, we can identify the direction ratios (or direction cosines) of the line as:
- \( a = 2 \)
- \( b = 3 \)
- \( c = -6 \)
### Step 3: Write the parametric equations of the line
Using the point \( P(1, -2, 3) \) and the direction ratios, we can write the parametric equations of the line:
\[
x = 1 + 2t
\]
\[
y = -2 + 3t
\]
\[
z = 3 - 6t
\]
where \( t \) is a parameter.
### Step 4: Substitute the parametric equations into the plane equation
We substitute \( x, y, z \) from the parametric equations into the plane equation \( x - y + z = 5 \):
\[
(1 + 2t) - (-2 + 3t) + (3 - 6t) = 5
\]
### Step 5: Simplify the equation
Now, simplifying the left-hand side:
\[
1 + 2t + 2 - 3t + 3 - 6t = 5
\]
Combining like terms:
\[
(1 + 2 + 3) + (2t - 3t - 6t) = 5
\]
\[
6 - 7t = 5
\]
### Step 6: Solve for \( t \)
Now, solve for \( t \):
\[
-7t = 5 - 6
\]
\[
-7t = -1
\]
\[
t = \frac{1}{7}
\]
### Step 7: Find the coordinates of the point on the plane
Now substitute \( t = \frac{1}{7} \) back into the parametric equations to find the coordinates of the point on the plane:
\[
x = 1 + 2\left(\frac{1}{7}\right) = 1 + \frac{2}{7} = \frac{9}{7}
\]
\[
y = -2 + 3\left(\frac{1}{7}\right) = -2 + \frac{3}{7} = -\frac{11}{7}
\]
\[
z = 3 - 6\left(\frac{1}{7}\right) = 3 - \frac{6}{7} = \frac{15}{7}
\]
### Step 8: Calculate the distance between the points
Now, we need to find the distance between the point \( P(1, -2, 3) \) and the point on the plane \( \left(\frac{9}{7}, -\frac{11}{7}, \frac{15}{7}\right) \) using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]
Substituting the values:
\[
d = \sqrt{\left(\frac{9}{7} - 1\right)^2 + \left(-\frac{11}{7} + 2\right)^2 + \left(\frac{15}{7} - 3\right)^2}
\]
Calculating each term:
\[
= \sqrt{\left(\frac{9}{7} - \frac{7}{7}\right)^2 + \left(-\frac{11}{7} + \frac{14}{7}\right)^2 + \left(\frac{15}{7} - \frac{21}{7}\right)^2}
\]
\[
= \sqrt{\left(\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(-\frac{6}{7}\right)^2}
\]
\[
= \sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}} = \sqrt{\frac{49}{49}} = \sqrt{1} = 1
\]
### Conclusion
Thus, the distance of the point \( (1, -2, 3) \) from the plane \( x - y + z = 5 \) measured parallel to the line is \( 1 \).
