<s>[INST] Finding the ratio of areas produced by perpendiculars from the $3$ sides of an equilateral triangle. 
A point O is inside an equilateral triangle $PQR$ and the perpendiculars $OL,OM,\text{and } ON$ are drawn to the sides $PQ,QR,\text{and } RP$ respectively. The ratios of lengths of the perpendiculars $OL:OM:ON \text{ is } 1:2:3$.
If  $\ \dfrac{\text{area of }LONP}{\text{area of }\Delta PQR}=\dfrac{a}{b}, \quad$ where $a$ 
  and $b$ are integers with no common factors,
what is the value of $a+b$ ?



All that I was able to do is:

Area $LONP=\frac{1}{2} |OL||PL|+\frac{1}{2} |NP||ON|=\frac{1}{2} |OL||PL|+\frac{1}{2} |NP|\ 3|OL|=\frac{1}{2} |OL|\ \left[\ |PL|+3|NP|\ \right]$
Area $PQR=\frac{1}{2} |PR||PQ|\sin 60^o=\frac{\sqrt 3}{4} |PR||PQ|=\frac{\sqrt 3}{4} |NP+RN||PL+LQ|=\frac{\sqrt 3}{4} \left[\ |NP|+|PL|+|RN|+|LQ| \ \right]$
Area $\Delta LON=\frac{1}{2} |ON||OL|\sin 120=\frac{\sqrt 3}{4} |OL|\ 3|OL|=\frac{3\sqrt 3}{4} |OL|^2$

$\mathbf{EDIT : }$Following Suraj M.S 's answer :

$$\begin{align}
\text{Area } \Delta PQR &=\dfrac{3x\ RN}{2}+\dfrac{3x\ PN}{2}+\dfrac{x\ PL}{2}+\dfrac{x\ QL}{2}+ {x\ QM}+{x\ MR} \\
\\
&=\dfrac{3x\ (PN+RN)}{2}+\dfrac{x\ (PL+QL)}{2}+{x\ (QM+MR)}\\
\\
&=\dfrac{3x\ (PR)}{2}+\dfrac{x\ (PQ)}{2}+{x\ (QR)}\\
\\
&=kx(\dfrac{3}{2}+\dfrac{1}{2}+{1})=3kx\\
\end{align}$$
Area $\Delta PQR=\dfrac{1}{2} k^2 \sin 60^o=\dfrac{k^2 \sqrt 3}{4} \implies x=\dfrac{k}{4\sqrt 3}$

$\mathbf{Question: }$How do I now find the area of $LONP$ in terms of $x'$s and/or $k'$s only ?

 [/INST] HINT:
Let $OL, OM, ON$ be $x, 2x, 3x$ respectively . Now you will observe 
$$area(\triangle PQR) = area(\triangle ONR) +area(\triangle ORM)+area(\triangle OMQ)+area(\triangle OQL)+area(\triangle OPL)+area(\triangle OPN)$$
since these are right angled and assuming $k$ be the side of the triangle you will get the resultant area as 
$$area(\triangle PQR)=3kx$$
equating it with the usual area
$$3kx=\frac{\sqrt{3}k^2}{2}$$
we get 
$$x=\frac{\sqrt{3}k}{6}$$
with the help of $x$ you will get $OL, OM, ON$.
Now let $\angle NPO=\theta$ then $\angle OPL=60^\circ-\theta$. with the help of the sides find $\frac{\sin \theta}{\sin (60^\circ-\theta)}$
you get
$$ \frac{\sin\theta}{\sin (60^\circ-\theta)}=\frac{ON}{OL}=3$$
solving for $\theta$
$$\tan \theta=\frac{3\sqrt{3}}{5}$$
with the help of $\tan$ you get $NP=\frac{5}{6}k$
now further you can get $PL$ by interchanging the angles $\theta$ and $60^\circ-\theta$ and once more solving  you get the new $\theta$ as 
$$\tan \theta=\frac{\sqrt{3}}{7}$$
using the same method of finding $NP$ find $PL$.
$$PL=\frac{7}{6}k$$
now we have found all our needed unknowns.
$$a=area(\triangle NOP)+area(\triangle POL)$$
$$=\frac{1}{2}(\frac{\sqrt{3}k}{2}.\frac{5k}{6}+ \frac{\sqrt{3}k}{6}.\frac{7k}{6} )$$
$$a=\frac{11\sqrt{3}k^2}{72}$$
also
$$b=\frac{\sqrt{3}k^2}{2}$$
solving you get $$\frac{a}{b}=\frac{11}{36}$$
$$a+b=\left(\frac{a}{b}+1\right)b$$
which solves to 
$$a+b=\frac{47\sqrt{3}k^2}{72}$$
where $k$ is the side of the equilateral triangle.
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