$1^{st}$ method: - equation of any straight line passing through the intersection of the lines x + 2y = 3 and 2x – 3y= 4 is

$\lambda$(x + 2y – 3) + (2x – 3y – 4) = 0

Since it passes through the point (2, 1)

$\lambda$(2 + 2 – 3) + (4 – 3 – 4) = 0

 $\lambda$ - 3 = 0

$\lambda$ = 3

Now substituting this value of $\lambda$  in (i), we get

3(x + 2y – 3) + (2x – 3y – 4) = 0

5x + 3y – 13 = 0

$2^{nd}$ method: - The straight line passing through the point (2, 1), put x = 2 and y = 1only option (a) and (b) will satisfy. Now, intersection point of the lines x + 2y = 3 and 2x – 3y= 4 is $(\frac{17}{7},\frac{2}{7}),$ Put x = $\frac{17}{7}$ and y = $\frac{2}{7}$ only option(a) will satisfy. Option(a).