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(C) The equation of any line passing through the point of intersection of $L = 0$ and $l = 0$ is given by $L + \lambda l = 0$.Substituting the given l

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(C) The equation of any line passing through the point of intersection of $L = 0$ and $l = 0$ is given by $L + \lambda l = 0$.Substituting the given lines:$(2x + 3y - 5) + \lambda(4x - 5y + 7) = 0$$(2 + 4\lambda)x + (3 - 5\lambda)y + (7\lambda - 5) = 0$  --- (Equation $1$)The line $L_1$ divides the segment joining $(2, 3)$ and $(1, -1)$ in the ratio $2:1$. Using the section formula,the point of division $(x, y)$ is:$x = \frac{2(1) + 1(2)}{2 + 1} = \frac{4}{3}$$y = \frac{2(-1) + 1(3)}{2 + 1} = \frac{1}{3}$Since the point $(\frac{4}{3}, \frac{1}{3})$ lies on $L_1$,we substitute it into Equation $1$:$(2 + 4\lambda)(\frac{4}{3}) + (3 - 5\lambda)(\frac{1}{3}) + (7\lambda - 5) = 0$Multiply by $3$ to clear the denominator:$4(2 + 4\lambda) + (3 - 5\lambda) + 3(7\lambda - 5) = 0$$8 + 16\lambda + 3 - 5\lambda + 21\lambda - 15 = 0$$32\lambda - 4 = 0 \Rightarrow \lambda = \frac{4}{32} = \frac{1}{8}$Substituting $\lambda = \frac{1}{8}$ back into Equation $1$:$(2 + 4(\frac{1}{8}))x + (3 - 5(\frac{1}{8}))y + (7(\frac{1}{8}) - 5) = 0$$(2 + 0.5)x + (3 - 0.625)y + (0.875 - 5) = 0$$2.5x + 2.375y - 4.125 = 0$$\frac{5}{2}x + \frac{19}{8}y = \frac{33}{8}$Multiply by $\frac{8}{33}$ to get the form $ax + by = 1$:$(\frac{5}{2} 	imes \frac{8}{33})x + (\frac{19}{8} 	imes \frac{8}{33})y = 1$$\frac{20}{33}x + \frac{19}{33}y = 1$Thus,$a = \frac{20}{33}$ and $b = \frac{19}{33}$.Therefore,$33(a - b) = 33(\frac{20}{33} - \frac{19}{33}) = 33(\frac{1}{33}) = 1$.

Let the line $L_1$ passing through the point of intersection of the lines $2x + 3y - 5 = 0$ and $4x - 5y + 7 = 0$ divide the line segment joining the points $(2, 3)$ and $(1, -1)$ in the ratio $2:1$. If the equation of $L_1$ is $ax + by = 1$,then $33(a - b) =$

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