Given the vectors $\vec{A}\, \vec{B}\, \vec{C}$. Find $\frac{mn}{p^2}$ if is known that $m\vec{A}+n\vec{B}+p\vec{C}=\vec{0}$.

$\begin{array}{ll}

1.&0.15\\

2.&0.21\\

3.&0.31\\

4.&1.25\\

5.&1.90\\

\end{array}$

I'm stuck on how to use the grid to get the coefficients on the given statement. Can this problem be solved graphically?.

So far the only thing which I could spot was that:

$\vec{A}=5\hat{i}+3\hat{j}$

$\vec{B}=2\hat{i}-5\hat{j}$

$\vec{C}=3\hat{i}-2\hat{j}$

Then by the information given:

$m\vec{A}+n\vec{B}+p\vec{C}=(5m+2n+3p)\hat{i}+(3m-5n-2p)\hat{j}=0$

From following the logic of the above equation:

$5m+2n+3p=0$

$3m-5n-2p=0$

But in this given situation there is no way to find all the unknowns as there's not enough equations to solve this system.

Assuming that I'm "eliminating p"

$5m+2n+3p=0$

$3m-5n-2p=0$

Multiplying by $2$ on the first equation and by $3$ on the second equation:

$10m+4n+6p=0$

$9m-15n-6p=0$

$19m-11n=0$

$m=\frac{11n}{19}$

Then for $p$:

$3(\frac{11n}{19})-5n=2p$

$p=\frac{-31n}{19}$

Then by returning to what it is being asked: Which it seems that it turns out that it is not necessary to know the third term in the equation.

$\frac{mn}{p^2}=\frac{11n}{19} \times n \times \left(\frac{19^2}{31^2n^2}\right)$

$\frac{mn}{p^2}=\frac{11n^2}{19} \times \left(\frac{19^2}{31^2n^2}\right)$

Then this results into:

$\frac{209}{961}$

which corresponds to $0.21748$ or the second option to which it does check with the given alternative. But does it exist an eassier method to solve this?.