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(C) Step $1$: Let the line be $y = 3 - x$. Substitute this into the equation of the pair of lines $x^2 - 3xy + y^2 = 0$.Step $2$: $x^2 - 3x(3 - x) + (

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$(\frac{3}{2}, \frac{3}{2})$

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(C) Step $1$: Let the line be $y = 3 - x$. Substitute this into the equation of the pair of lines $x^2 - 3xy + y^2 = 0$.Step $2$: $x^2 - 3x(3 - x) + (3 - x)^2 = 0$.Step $3$: $x^2 - 9x + 3x^2 + 9 - 6x + x^2 = 0$.Step $4$: $5x^2 - 15x + 9 = 0$. Let the roots be $x_1$ and $x_2$, which are the $x$-coordinates of $A$ and $B$.Step $5$: The $x$-coordinate of the midpoint is $\frac{x_1 + x_2}{2} = \frac{-(-15)}{2 	imes 5} = \frac{15}{10} = \frac{3}{2}$.Step $6$: Since the midpoint lies on the line $x + y = 3$, the $y$-coordinate is $y = 3 - \frac{3}{2} = \frac{3}{2}$.Step $7$: Thus, the midpoint is $(\frac{3}{2}, \frac{3}{2})$.

The line $x + y = 3$ intersects the pair of straight lines $x^2 - 3xy + y^2 = 0$ at points $A$ and $B$. Then the coordinates of the mid-point of $AB$ are

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