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(D) The equation of the family of lines passing through the intersection of $x+2y-19=0$ and $x-2y-3=0$ is given by $(x+2y-19) + \lambda(x-2y-3) = 0$.T

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-$14$

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(D) The equation of the family of lines passing through the intersection of $x+2y-19=0$ and $x-2y-3=0$ is given by $(x+2y-19) + \lambda(x-2y-3) = 0$.This simplifies to $(1+\lambda)x + (2-2\lambda)y - (19+3\lambda) = 0$.The perpendicular distance from the point $(-2,4)$ to this line is $5$ units.Using the distance formula $d = \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$,we have $5 = \frac{|(1+\lambda)(-2) + (2-2\lambda)(4) - (19+3\lambda)|}{\sqrt{(1+\lambda)^2 + (2-2\lambda)^2}}$.$5 = \frac{|-2-2\lambda + 8-8\lambda - 19-3\lambda|}{\sqrt{1+2\lambda+\lambda^2 + 4-8\lambda+4\lambda^2}}$.$5 = \frac{|-13\lambda - 13|}{\sqrt{5\lambda^2-6\lambda+5}}$.Squaring both sides: $25(5\lambda^2-6\lambda+5) = (-13\lambda-13)^2$.$125\lambda^2 - 150\lambda + 125 = 169\lambda^2 + 338\lambda + 169$.$44\lambda^2 + 488\lambda + 44 = 0 \Rightarrow 11\lambda^2 + 122\lambda + 11 = 0$.$(11\lambda+1)(\lambda+11) = 0$,so $\lambda = -1/11$ or $\lambda = -11$.For $\lambda = -1/11$: $(1-1/11)x + (2+2/11)y - (19-3/11) = 0$ $\Rightarrow (10/11)x + (24/11)y - (206/11) = 0$ $\Rightarrow 5x+12y-103=0$.Here $b=12, c=-103$,so $5+b+c = 5+12-103 = -86$.For $\lambda = -11$: $(1-11)x + (2+22)y - (19-33) = 0$ $\Rightarrow -10x+24y+14=0$ $\Rightarrow 5x-12y-7=0$.Here $b=-12, c=-7$,so $5+b+c = 5-12-7 = -14$.

If the equation of the straight line passing through the point of intersection of $x+2y-19=0$ and $x-2y-3=0$ and which is at a perpendicular distance of $5$ units from the point $(-2,4)$ is $5x+by+c=0$,then a possible value of $5+b+c$ is

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