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

We have the fixed points $$A(1,-1)$$ and $$B(0,2)$$ and a variable point $$P(x',y')$$ that lies on the straight line

$$3x+y-4\lambda = 0.$$

Therefore every such point must satisfy

$$3x' + y' = 4\lambda. \quad -(1)$$

The area of the triangle formed by the three points is given to be $$5$$ square units. For three points $$P(x_1,y_1),\;A(x_2,y_2),\;B(x_3,y_3)$$ the area formula is

$$\text{Area}=\frac12\left|\,x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\,\right|.$$

Putting $$P(x',y'),\;A(1,-1),\;B(0,2)$$ into the formula, we get

$$\text{Area}=\frac12\left|\,x'( -1-2) + 1(2-y') + 0(y' +1)\,\right|.$$

Simplifying the expression inside the modulus step by step,

$$x'( -1-2)=x'(-3)=-3x',$$ $$1(2-y')=2-y',$$ and the last term is zero. Hence

$$\text{Area}=\frac12\left|\, -3x' + 2 - y'\,\right|.$$

Given that this area equals $$5$$, we write

$$\frac12\left| -3x' + 2 - y' \right| = 5.$$

Multiplying by $$2$$ gives

$$\left| -3x' + 2 - y' \right| = 10.$$

Because a modulus is unchanged by an overall sign, we may reverse the order to obtain

$$\left|\,3x' + y' - 2\,\right| = 10. \quad -(2)$$

Now we substitute the line condition (1), namely $$3x'+y'=4\lambda,$$ into (2):

$$\left|\,4\lambda - 2\,\right| = 10.$$

Removing the absolute value leads to the two equations

$$4\lambda - 2 = 10 \quad\text{or}\quad 4\lambda - 2 = -10.$$

Solving each, we get

First case: $$4\lambda = 12 \;\Rightarrow\; \lambda = 3.$$ Second case: $$4\lambda = -8 \;\Rightarrow\; \lambda = -2.$$

Among the answer choices only $$\lambda = 3$$ is listed.

Hence, the correct answer is Option B.