To solve the problem, we need to find the ratio in which the plane \(2x - 3y + z = 8\) divides the line segment joining the points \(A(3, -2, 1)\) and \(B(1, 4, -3)\). We will also find the point of intersection of the line and the plane.
### Step 1: Determine the coordinates of the point dividing the line segment in the ratio \(m:n\)
Let the point of intersection be \(P(\alpha, \beta, \gamma)\). According to the section formula, if a point divides the line segment joining points \(A(x_1, y_1, z_1)\) and \(B(x_2, y_2, z_2)\) in the ratio \(m:n\), then the coordinates of point \(P\) can be expressed as:
\[
\alpha = \frac{mx_2 + nx_1}{m+n}, \quad \beta = \frac{my_2 + ny_1}{m+n}, \quad \gamma = \frac{mz_2 + nz_1}{m+n}
\]
For our case, \(A(3, -2, 1)\) and \(B(1, 4, -3)\). Let the ratio in which the plane divides the line segment be \(m:1\) (where \(n = 1\)).
Thus, we have:
\[
\alpha = \frac{m \cdot 1 + 1 \cdot 3}{m + 1}, \quad \beta = \frac{m \cdot 4 + 1 \cdot (-2)}{m + 1}, \quad \gamma = \frac{m \cdot (-3) + 1 \cdot 1}{m + 1}
\]
### Step 2: Substitute the coordinates into the plane equation
The coordinates \((\alpha, \beta, \gamma)\) must satisfy the equation of the plane \(2x - 3y + z = 8\). Therefore, we substitute \(\alpha\), \(\beta\), and \(\gamma\) into the plane equation:
\[
2\left(\frac{m \cdot 1 + 3}{m + 1}\right) - 3\left(\frac{4m - 2}{m + 1}\right) + \left(\frac{-3m + 1}{m + 1}\right) = 8
\]
### Step 3: Simplify the equation
Now, we simplify the left-hand side:
\[
\frac{2(m + 3) - 3(4m - 2) - 3m + 1}{m + 1} = 8
\]
This simplifies to:
\[
\frac{2m + 6 - 12m + 6 - 3m + 1}{m + 1} = 8
\]
Combining like terms gives:
\[
\frac{-13m + 13}{m + 1} = 8
\]
### Step 4: Cross-multiply and solve for \(m\)
Cross-multiplying gives:
\[
-13m + 13 = 8(m + 1)
\]
Expanding the right side:
\[
-13m + 13 = 8m + 8
\]
Rearranging terms:
\[
13 - 8 = 8m + 13m
\]
\[
5 = 21m
\]
Thus, we find:
\[
m = \frac{5}{21}
\]
### Step 5: Determine the ratio
The ratio in which the plane divides the line segment \(AB\) is \(m:1 = \frac{5}{21}:1\). Therefore, the ratio is:
\[
5:21
\]
### Step 6: Find the coordinates of the point of intersection
Now, we substitute \(m = \frac{5}{21}\) back into the expressions for \(\alpha\), \(\beta\), and \(\gamma\):
\[
\alpha = \frac{\frac{5}{21} \cdot 1 + 3}{\frac{5}{21} + 1} = \frac{\frac{5}{21} + \frac{63}{21}}{\frac{5}{21} + \frac{21}{21}} = \frac{\frac{68}{21}}{\frac{26}{21}} = \frac{68}{26} = \frac{34}{13}
\]
\[
\beta = \frac{\frac{5}{21} \cdot 4 - 2}{\frac{5}{21} + 1} = \frac{\frac{20}{21} - \frac{42}{21}}{\frac{26}{21}} = \frac{-\frac{22}{21}}{\frac{26}{21}} = -\frac{22}{26} = -\frac{11}{13}
\]
\[
\gamma = \frac{\frac{5}{21} \cdot (-3) + 1}{\frac{5}{21} + 1} = \frac{-\frac{15}{21} + \frac{21}{21}}{\frac{26}{21}} = \frac{\frac{6}{21}}{\frac{26}{21}} = \frac{6}{26} = \frac{3}{13}
\]
### Final Result
Thus, the coordinates of the point of intersection are:
\[
P\left(\frac{34}{13}, -\frac{11}{13}, \frac{3}{13}\right)
\]
### Summary of Results
1. The ratio in which the plane divides the line segment \(AB\) is \(5:21\).
2. The point of intersection of the line and the plane is \(P\left(\frac{34}{13}, -\frac{11}{13}, \frac{3}{13}\right)\).
