1. Given points

$A(-1,1),\quad B(3,4),\quad C(2,0)$

A line through the origin is

$y=mx, \quad m>0$

It intersects line $AC$ at $P$ and line $BC$ at $Q$.

We are given

$A_1=[\triangle ABC], \quad A_2=[\triangle PQC], \quad A_1=3A_2$

So,

$[\triangle PQC]=\frac13[\triangle ABC]$

2. Equation of line $AC$

Slope of $AC$:

$\frac{0-1}{2-(-1)}=-\frac13$

So equation through $A(-1,1)$:

$y-1=-\frac13(x+1)$

$3y-3=-x-1$

$x+3y-2=0$

Hence,

$AC: x+3y-2=0$

3. Equation of line $BC$

Slope of $BC$:

$\frac{0-4}{2-3}=4$

Equation through $C(2,0)$:

$y=4(x-2)$

$y=4x-8$

Hence,

$BC: 4x-y-8=0$

4. Coordinates of $P$

$P$ lies on both $AC$ and $y=mx$.

Substitute $y=mx$ into $x+3y-2=0$:

$x+3mx-2=0$

$x(1+3m)=2$

$x_P=\frac{2}{1+3m}$

$y_P=m x_P=\frac{2m}{1+3m}$

Thus,

$P\left(\frac{2}{1+3m},\frac{2m}{1+3m}\right)$

5. Coordinates of $Q$

$Q$ lies on both $BC$ and $y=mx$.

Substitute $y=mx$ into $y=4x-8$:

$mx=4x-8$

$(4-m)x=8$

$x_Q=\frac{8}{4-m}$

$y_Q=m x_Q=\frac{8m}{4-m}$

Thus,

$Q\left(\frac{8}{4-m},\frac{8m}{4-m}\right)$

6. Use area ratio via similarity

Since $P$ lies on $AC$, $Q$ lies on $BC$, and $PQ$ is along the same direction as line through origin, triangle $PQC$ is formed inside triangle $ABC$ with vertex $C$ and sides along $CA$ and $CB$.

Let us parametrize points on the two sides from $C$.

For point $P$ on $AC$:

$A-C=(-1-2,1-0)=(-3,1)$

So any point on $AC$ is

$C+t(A-C)=(2-3t,t)$

Comparing with $P$:

$y_P=t=\frac{2m}{1+3m}$

Hence,

$\frac{CP}{CA}=t=\frac{2m}{1+3m}$

For point $Q$ on $BC$:

$B-C=(3-2,4-0)=(1,4)$

So any point on $BC$ is

$C+s(B-C)=(2+s,4s)$

Comparing with $Q$:

$4s=\frac{8m}{4-m}$

$s=\frac{2m}{4-m}$

Hence,

$\frac{CQ}{CB}=s=\frac{2m}{4-m}$

Now,

$[\triangle PQC]=\frac{CP}{CA}\cdot \frac{CQ}{CB}\cdot [\triangle ABC]$

Therefore,

$\frac{A_2}{A_1}=\frac{2m}{1+3m}\cdot\frac{2m}{4-m}$

Given $A_1=3A_2$, so

$\frac{A_2}{A_1}=\frac13$

Thus,

$\frac{2m}{1+3m}\cdot\frac{2m}{4-m}=\frac13$

7. Solve for $m$

$\frac{4m^2}{(1+3m)(4-m)}=\frac13$

$12m^2=(1+3m)(4-m)$

Expand RHS:

$(1+3m)(4-m)=4-m+12m-3m^2=4+11m-3m^2$

So,

$12m^2=4+11m-3m^2$

$15m^2-11m-4=0$

Factorize:

$15m^2-15m+4m-4=0$

$15m(m-1)+4(m-1)=0$

$(m-1)(15m+4)=0$

So,

$m=1 \quad \text{or} \quad m=-\frac{4}{15}$

Given $m>0$, hence

$m=1$

8. Check with options

Option A: $1$ ✅

So the correct answer is

$\boxed{1}$

9. Comparison with stored correct answer

Stored correct answer: A

Our derived answer is also Option A, i.e. $m=1$.

So the stored answer is correct.