I don't know why its not easy...

There are many similar triangles in this diagram: $\triangle SUP$ , $\triangle TRU$ , and $\triangle QRP$ are all similar to each other, thanks to the two pairs of parallel lines.

Now the ratio of the areas of two similar triangles is the square of the ratio of the side lengths of the corresponding sides. Let $A$ denote the areas of the various shapes. Thus, since $A (\triangle SUP) = \frac{4}{9} A( \triangle TRU)$ , it follows that $UP = \frac{2}{3} RU$ . Thus $UP = \frac{2}{5} RP,$ and $120 = A( \triangle SUP) = \frac{4}{25} A( \triangle QRP)$ , and $A (\triangle QRP) = 750$ .

After subtracting the areas of $\triangle SUP$ and $\triangle TRU$ , we see that the area of the parallelogram $SQTU$ is $750 - 120 - 270 = 360$ . The two triangles $\triangle SQT$ and $\triangle TUS$ are congruent and thus the area of each is $180$ .

@Syed Hamza Khalid solution of "https://brilliant.org/discussions/thread/how-to-approach-this-question/?from_notification=14176037#comment-a3f5142823e92"

We note that $\triangle PSU$ , $\triangle TUR$ , and $\triangle PQR$ are similar. Then

$\begin{aligned} \frac {UR^2}{PU^2} & = \frac {270}{120} = \frac 94 \\ \implies \frac {UR}{PU} & = \frac 32 \end{aligned}$

Now, we have:

$\begin{aligned} \frac {[PQR]}{[PSU]} & = \frac {PR^2}{PU^2} = \left(\frac {PU+UR}{PU}\right) = \left(1+\frac 32 \right)^2 = \frac {25}4 & \small \color{#3D99F6} [\cdots ] \text{ denotes the area of }\cdots. \\ \implies [PQR] & = \frac {25}4 [PSU] = \frac {25}4 \times 120 = 750 \end{aligned}$

we note that:

$\begin{aligned} [QST] & = \frac {[QTUS]}2 = \frac {[PQR]-[PSU]-[TUR]}2 = \frac {750-120-270}2 = \boxed{180}\end{aligned}$