Total price of $6\text{ shirts}$ and $7\text{ trousers}$ is $\text{Rs. }4,130$ while that of $4\text{ shirts}$ and $9\text{ trousers}$ is $\text{Rs. }4,270$. Find out the price of $3\text{ shirts}$.

Let the cost of one shirt be $S$ and the cost of one pair of trousers be $T$.
From the given details, write a pair of simultaneous linear equations:

$$6S+7T=4130\text{— (Equation 1)}$$

$$4S+9T=4270\text{— (Equation 2)}$$

To eliminate $S$, multiply Equation 1 by $2$ and Equation 2 by $3$:

$$12S+14T=8260\text{— (Equation 3)}$$

$$12S+27T=12810\text{— (Equation 4)}$$

Subtract Equation 3 from Equation 4:

$$(12S+27T)-(12S+14T)=12810-8260$$

$$13T=4550$$

$$T=\frac{4550}{13}=350$$

Substitute the value of $T=350$ back into Equation 1:

$$6S+7(350)=4130$$

$$6S+2450=4130$$

$$6S=4130-2450$$

$$6S=1680$$

$$S=\frac{1680}{6}=280$$

The cost of one shirt is $\text{Rs. }280$. Therefore, the cost of $3\text{ shirts}$ is:

$$\text{Cost of 3 shirts}=3\times 280=\text{Rs. }840$$