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(C) Let the intercepts be $5C$ and $3C$. The equation is $\frac{x}{5C} + \frac{y}{3C} = 1$. Since it passes through $(-4, 3)$,we have $-\frac{4}{5C} +

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$A-2, B-5, C-3, D-1$

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(C) Let the intercepts be $5C$ and $3C$. The equation is $\frac{x}{5C} + \frac{y}{3C} = 1$. Since it passes through $(-4, 3)$,we have $-\frac{4}{5C} + \frac{3}{3C} = 1$ $\Rightarrow \frac{-4+5}{5C} = 1$ $\Rightarrow 5C = 1$ $\Rightarrow C = \frac{1}{5}$. Thus,$\frac{x}{1} + \frac{y}{3/5} = 1$ $\Rightarrow x + \frac{5y}{3} = 1$ $\Rightarrow 3x + 5y = 3$. So,$A-2$.$(B)$ Let the line be $\frac{x}{a} + \frac{y}{b} = 1$. The intercepts are $A(a, 0)$ and $B(0, b)$. $P(2, -5)$ is the midpoint,so $\frac{a}{2} = 2 \Rightarrow a = 4$ and $\frac{b}{2} = -5 \Rightarrow b = -10$. The equation is $\frac{x}{4} + \frac{y}{-10} = 1$ $\Rightarrow 5x - 2y = 20$ $\Rightarrow 5x - 2y - 20 = 0$. So,$B-5$.$(C)$ Line parallel to $2x - 3y + 5 = 0$ is $2x - 3y + k = 0$. It has $x$-intercept $\frac{2}{5}$,so at $y=0, x=\frac{2}{5}$ $\Rightarrow 2(\frac{2}{5}) - 3(0) + k = 0$ $\Rightarrow k = -\frac{4}{5}$. The equation is $2x - 3y - \frac{4}{5} = 0$ $\Rightarrow 10x - 15y - 4 = 0$ $\Rightarrow 10x - 15y = 4$. So,$C-4$.$(D)$ Line perpendicular to $5x + 2y + 7 = 0$ is $2x - 5y + \mu = 0$. It has $y$-intercept $\frac{4}{5}$,so at $x=0, y=\frac{4}{5}$ $\Rightarrow 2(0) - 5(\frac{4}{5}) + \mu = 0$ $\Rightarrow -4 + \mu = 0$ $\Rightarrow \mu = 4$. The equation is $2x - 5y + 4 = 0$. So,$D-1$.

List-$I$	List-$II$
$A$. Line passing through $(-4, 3)$ and having intercepts in the ratio $5:3$	$1$. $2x - 5y + 4 = 0$
$B$. Line passing through $P(2, -5)$ such that $P$ bisects the part intercepted between the axes	$2$. $3x + 5y = 3$
$C$. Line parallel to $2x - 3y + 5 = 0$ with $x$-intercept $\frac{2}{5}$ is	$3$. $10x - 15y + 4 = 0$
$D$. Line perpendicular to $5x + 2y + 7 = 0$ with $y$-intercept $\frac{4}{5}$ is	$4$. $10x - 15y = 4$
	$5$. $5x - 2y - 20 = 0$

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