Let two points be `A(1,-1)` and `B(0, 2)`. If a point P(x', y') be such that the area of `DeltaPAB = 5` sq. units and it lies on the line, `3x + y - 4lambda =0`, then a value of `lambda` is:

Options

A. 4
B. 1
C. `-3`
D. 3

Answer

To solve the problem, we need to find the value of \( \lambda \) such that the area of triangle \( PAB \) is 5 square units, where \( A(1, -1) \), \( B(0, 2) \), and \( P(x', y') \) lies on the line \( 3x + y - 4\lambda = 0 \).

### Step-by-Step Solution:

1. **Area of Triangle Formula**:
The area of triangle formed by points \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( P(x_3, y_3) \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For our points \( A(1, -1) \), \( B(0, 2) \), and \( P(x', y') \), we substitute:
\[ \text{Area} = \frac{1}{2} \left| 1(2 - y') + 0(y' + 1) + x'(-1 - 2) \right| \]
Simplifying this gives:
\[ \text{Area} = \frac{1}{2} \left| 2 - y' - 3x' \right| \]

2. **Setting the Area**:
Since the area is given as 5 square units, we set up the equation:
\[ \frac{1}{2} \left| 2 - y' - 3x' \right| = 5 \]
Multiplying both sides by 2:
\[ \left| 2 - y' - 3x' \right| = 10 \]

3. **Removing the Absolute Value**:
This gives us two cases to consider:
- Case 1: \( 2 - y' - 3x' = 10 \)
- Case 2: \( 2 - y' - 3x' = -10 \)

**Case 1**:
\[ -y' - 3x' = 8 \implies y' + 3x' = -8 \quad \text{(Equation 1)} \]

**Case 2**:
\[ -y' - 3x' = -12 \implies y' + 3x' = 12 \quad \text{(Equation 2)} \]

4. **Substituting into the Line Equation**:
The point \( P(x', y') \) lies on the line \( 3x + y - 4\lambda = 0 \). Therefore, we can substitute \( y' \) from both equations into the line equation.

**From Equation 1**:
\[ y' = -8 - 3x' \]
Substituting into the line equation:
\[ 3x' + (-8 - 3x') - 4\lambda = 0 \implies -8 - 4\lambda = 0 \implies 4\lambda = -8 \implies \lambda = -2 \]

**From Equation 2**:
\[ y' = 12 - 3x' \]
Substituting into the line equation:
\[ 3x' + (12 - 3x') - 4\lambda = 0 \implies 12 - 4\lambda = 0 \implies 4\lambda = 12 \implies \lambda = 3 \]

5. **Conclusion**:
We have two possible values for \( \lambda \):
- \( \lambda = -2 \)
- \( \lambda = 3 \)

Since we need a value of \( \lambda \), we can choose \( \lambda = 3 \) as the valid solution.

### Final Answer:
The value of \( \lambda \) is \( 3 \).