To solve the problem, we need to find the area of triangle ABC formed by the points A, B, and C, where:
- A is given as \( A(3, -4, 1) \).
- B is the intersection of the line \( L \) with the plane \( P \).
- C is the intersection of the line \( L \) with the xy-plane.
### Step 1: Find the coordinates of point B
The line \( L \) is given in symmetric form:
\[
\frac{x-3}{2} = \frac{y+1}{-3} = \frac{z-2}{-1} = \lambda
\]
From this, we can express the coordinates \( x, y, z \) in terms of \( \lambda \):
\[
x = 2\lambda + 3, \quad y = -3\lambda - 1, \quad z = 2 - \lambda
\]
Now, substitute these expressions into the plane equation \( P: 2x + y - z = 5 \):
\[
2(2\lambda + 3) + (-3\lambda - 1) - (2 - \lambda) = 5
\]
Expanding this gives:
\[
4\lambda + 6 - 3\lambda - 1 - 2 + \lambda = 5
\]
Combining like terms:
\[
(4\lambda - 3\lambda + \lambda) + (6 - 1 - 2) = 5
\]
\[
2\lambda + 3 = 5
\]
Solving for \( \lambda \):
\[
2\lambda = 2 \implies \lambda = 1
\]
Now substitute \( \lambda = 1 \) back into the equations for \( x, y, z \):
\[
x = 2(1) + 3 = 5, \quad y = -3(1) - 1 = -4, \quad z = 2 - 1 = 1
\]
Thus, the coordinates of point B are \( B(5, -4, 1) \).
### Step 2: Find the coordinates of point C
Point C is the intersection of line \( L \) with the xy-plane, where \( z = 0 \):
Set \( z = 0 \) in the equation for \( z \):
\[
2 - \lambda = 0 \implies \lambda = 2
\]
Now substitute \( \lambda = 2 \) back into the equations for \( x, y \):
\[
x = 2(2) + 3 = 7, \quad y = -3(2) - 1 = -7
\]
Thus, the coordinates of point C are \( C(7, -7, 0) \).
### Step 3: Calculate the area of triangle ABC
The area of triangle ABC can be calculated using the formula:
\[
\text{Area} = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right|
\]
First, find the vectors \( \vec{AB} \) and \( \vec{AC} \):
\[
\vec{AB} = B - A = (5 - 3, -4 + 4, 1 - 1) = (2, 0, 0)
\]
\[
\vec{AC} = C - A = (7 - 3, -7 + 4, 0 - 1) = (4, -3, -1)
\]
Now compute the cross product \( \vec{AB} \times \vec{AC} \):
\[
\vec{AB} \times \vec{AC} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 0 & 0 \\
4 & -3 & -1
\end{vmatrix}
\]
Calculating the determinant:
\[
= \hat{i} \begin{vmatrix} 0 & 0 \\ -3 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 0 \\ 4 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 0 \\ 4 & -3 \end{vmatrix}
\]
Calculating each of these determinants:
1. \( \hat{i}(0 \cdot -1 - 0 \cdot -3) = 0 \)
2. \( -\hat{j}(2 \cdot -1 - 0 \cdot 4) = -(-2) = 2 \)
3. \( \hat{k}(2 \cdot -3 - 0 \cdot 4) = -6 \)
So,
\[
\vec{AB} \times \vec{AC} = (0, 2, -6)
\]
Now find the magnitude:
\[
\left| \vec{AB} \times \vec{AC} \right| = \sqrt{0^2 + 2^2 + (-6)^2} = \sqrt{0 + 4 + 36} = \sqrt{40} = 2\sqrt{10}
\]
Finally, the area of triangle ABC is:
\[
\text{Area} = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right| = \frac{1}{2} \cdot 2\sqrt{10} = \sqrt{10}
\]
### Final Answer
The area of triangle ABC is \( \sqrt{10} \) square units.
