1. Equation of line through $A(1,-1)$ and $B(0,2)$

Slope of $AB$ is $m=\frac{2-(-1)}{0-1}=\frac{3}{-1}=-3.$

So the equation of line $AB$ is $y+1=-3(x-1)$ $y=-3x+2$ $3x+y-2=0.$

2. Use area formula for triangle $PAB$

Given area of $\triangle PAB = 5$.

If $P(x',y')$, then using base $AB$, area can be written as $\text{Area} = \frac{1}{2}\times (\text{length of }AB)\times (\text{distance of }P\text{ from line }AB).$

First find $AB$: $AB=\sqrt{(1-0)^2+(-1-2)^2}=\sqrt{1+9}=\sqrt{10}.$

Distance of $P(x',y')$ from line $3x+y-2=0$ is $d=\frac{|3x'+y'-2|}{\sqrt{3^2+1^2}}=\frac{|3x'+y'-2|}{\sqrt{10}}.$

Hence, $5=\frac{1}{2}\cdot \sqrt{10}\cdot \frac{|3x'+y'-2|}{\sqrt{10}}$ $5=\frac{1}{2}|3x'+y'-2|$ $|3x'+y'-2|=10.$

So, $3x'+y'-2=\pm 10$ which gives $3x'+y'=12 \quad \text{or} \quad 3x'+y'=-8.$

Thus point $P$ must lie on one of the two lines: $3x+y-12=0 \quad \text{or} \quad 3x+y+8=0.$

3. Given that $P$ lies on the line $3x+y-4\lambda=0.$

For this line to be one of the above two lines, we must have $-4\lambda=-12 \quad \text{or} \quad -4\lambda=8.$

So, $\lambda=3 \quad \text{or} \quad \lambda=-2.$

4. Match with options

Options are:

• A: $4$
• B: $1$
• C: $-3$
• D: $3$

Among these, only $\lambda=3$ is present.

So the correct option is D.