To find the ratio in which the plane \(2x + 3y + 5z = 1\) divides the line segment joining the points \(P(1, 0, -3)\) and \(Q(1, -5, 7)\), we can follow these steps:
### Step 1: Determine the coordinates of the points
We have two points:
- \(P(1, 0, -3)\)
- \(Q(1, -5, 7)\)
### Step 2: Use the section formula
Let the point \(A\) divide the line segment \(PQ\) in the ratio \(k:1\). According to the section formula, the coordinates of point \(A\) can be calculated as:
\[
A\left(\frac{k \cdot x_2 + 1 \cdot x_1}{k + 1}, \frac{k \cdot y_2 + 1 \cdot y_1}{k + 1}, \frac{k \cdot z_2 + 1 \cdot z_1}{k + 1}\right)
\]
Substituting the coordinates of points \(P\) and \(Q\):
- \(x_1 = 1, y_1 = 0, z_1 = -3\)
- \(x_2 = 1, y_2 = -5, z_2 = 7\)
The coordinates of point \(A\) become:
\[
A\left(\frac{k \cdot 1 + 1 \cdot 1}{k + 1}, \frac{k \cdot (-5) + 1 \cdot 0}{k + 1}, \frac{k \cdot 7 + 1 \cdot (-3)}{k + 1}\right)
\]
This simplifies to:
\[
A\left(1, \frac{-5k}{k + 1}, \frac{7k - 3}{k + 1}\right)
\]
### Step 3: Substitute point \(A\) into the plane equation
Since point \(A\) lies on the plane \(2x + 3y + 5z = 1\), we substitute the coordinates of \(A\) into the plane equation:
\[
2(1) + 3\left(\frac{-5k}{k + 1}\right) + 5\left(\frac{7k - 3}{k + 1}\right) = 1
\]
This simplifies to:
\[
2 - \frac{15k}{k + 1} + \frac{35k - 15}{k + 1} = 1
\]
Combining the terms gives:
\[
2 + \frac{20k - 15}{k + 1} = 1
\]
### Step 4: Clear the fraction
Multiply through by \(k + 1\) to eliminate the denominator:
\[
2(k + 1) + 20k - 15 = k + 1
\]
Expanding and simplifying:
\[
2k + 2 + 20k - 15 = k + 1
\]
\[
22k - k + 2 - 15 = 1
\]
\[
21k - 13 = 1
\]
\[
21k = 14 \implies k = \frac{14}{21} = \frac{2}{3}
\]
### Step 5: Find the ratio
The ratio \(k:1\) is:
\[
\frac{2}{3}:1 = 2:3
\]
### Final Answer
Thus, the ratio in which the plane divides the line segment joining the points \(P\) and \(Q\) is \(2:3\).
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