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(C) The equation of a line passing through the intersection of $L_1: x + y - 3 = 0$ and $L_2: 2x - y + 1 = 0$ is given by $(x + y - 3) + \lambda(2x -

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$4x + y - 5 = 0$

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(C) The equation of a line passing through the intersection of $L_1: x + y - 3 = 0$ and $L_2: 2x - y + 1 = 0$ is given by $(x + y - 3) + \lambda(2x - y + 1) = 0$.Since this line passes through the point $(2, -3)$,we substitute $x = 2$ and $y = -3$ into the equation:$(2 - 3 - 3) + \lambda(2(2) - (-3) + 1) = 0$$-4 + \lambda(4 + 3 + 1) = 0$$-4 + 8\lambda = 0$$8\lambda = 4 \implies \lambda = \frac{1}{2}$.Substituting $\lambda = \frac{1}{2}$ back into the family of lines equation:$(x + y - 3) + \frac{1}{2}(2x - y + 1) = 0$Multiply by $2$ to clear the fraction:$2(x + y - 3) + (2x - y + 1) = 0$$2x + 2y - 6 + 2x - y + 1 = 0$$4x + y - 5 = 0$.

Find the equation of the line passing through the intersection of the lines $x + y - 3 = 0$ and $2x - y + 1 = 0$ and the point $(2, -3)$.

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