To solve the problem, we need to find the values of \(\lambda\) given the triangle \(ABC\) with vertices \(A(1, -1)\), \(B(0, 2)\), and \(C(x', y')\) where the area of triangle \(ABC\) is 5, and point \(C\) lies on the line defined by the equation \(3x + y - 4\lambda = 0\).
### Step 1: Use the Area Formula for a Triangle
The area \(A\) of triangle \(ABC\) can be calculated using the formula:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Where \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) are the coordinates of points \(A\), \(B\), and \(C\) respectively.
### Step 2: Substitute the Coordinates
Substituting the coordinates of points \(A\), \(B\), and \(C\):
- \(A(1, -1)\) \(\Rightarrow x_1 = 1, y_1 = -1\)
- \(B(0, 2)\) \(\Rightarrow x_2 = 0, y_2 = 2\)
- \(C(x', y')\) \(\Rightarrow x_3 = x', y_3 = y'\)
The area becomes:
\[
5 = \frac{1}{2} \left| 1(2 - y') + 0(y' + 1) + x'(-1 - 2) \right|
\]
This simplifies to:
\[
5 = \frac{1}{2} \left| 2 - y' - 3x' \right|
\]
### Step 3: Remove the Absolute Value
Removing the absolute value gives us two equations:
\[
2 - y' - 3x' = 10 \quad \text{(1)}
\]
\[
2 - y' - 3x' = -10 \quad \text{(2)}
\]
### Step 4: Solve the Equations
From equation (1):
\[
-3x' - y' = 8 \quad \Rightarrow 3x' + y' = -8 \quad \text{(3)}
\]
From equation (2):
\[
-3x' - y' = -12 \quad \Rightarrow 3x' + y' = 2 \quad \text{(4)}
\]
### Step 5: Use the Line Equation
The point \(C(x', y')\) lies on the line given by:
\[
3x' + y' - 4\lambda = 0
\]
Substituting equations (3) and (4) into this line equation:
1. From equation (3):
\[
-4\lambda = -8 \quad \Rightarrow \lambda = 2
\]
2. From equation (4):
\[
-4\lambda = 2 \quad \Rightarrow \lambda = -\frac{1}{2}
\]
### Step 6: Final Values of \(\lambda\)
Thus, the possible values of \(\lambda\) are \(2\) and \(-\frac{1}{2}\).
### Conclusion
The values of \(\lambda\) are:
\[
\lambda = 2 \quad \text{and} \quad \lambda = -\frac{1}{2}
\]
