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(C) Given the line $3x + 4y = 12$. Let the coordinates of $P$ be $(r \cos 	heta, r \sin 	heta)$ and $Q$ be $(r \cos(90^{\circ} + 	heta), r \sin(90^

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$46$

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(C) Given the line $3x + 4y = 12$. Let the coordinates of $P$ be $(r \cos 	heta, r \sin 	heta)$ and $Q$ be $(r \cos(90^{\circ} + 	heta), r \sin(90^{\circ} + 	heta)) = (-r \sin 	heta, r \cos 	heta)$ since $	riangle OPQ$ is isosceles with $OP = OQ = r$ and $\angle POQ = 90^{\circ}$.Since $P$ and $Q$ lie on the line $3x + 4y = 12$:For $P$: $3(r \cos 	heta) + 4(r \sin 	heta) = 12 \Rightarrow r(3 \cos 	heta + 4 \sin 	heta) = 12 \ldots(1)$For $Q$: $3(-r \sin 	heta) + 4(r \cos 	heta) = 12 \Rightarrow r(4 \cos 	heta - 3 \sin 	heta) = 12 \ldots(2)$Squaring and adding $(1)$ and $(2)$:$r^2(3 \cos 	heta + 4 \sin 	heta)^2 + r^2(4 \cos 	heta - 3 \sin 	heta)^2 = 12^2 + 12^2$$r^2(9 \cos^2 	heta + 16 \sin^2 	heta + 24 \sin 	heta \cos 	heta + 16 \cos^2 	heta + 9 \sin^2 	heta - 24 \sin 	heta \cos 	heta) = 288$$r^2(25 \cos^2 	heta + 25 \sin^2 	heta) = 288$ $\Rightarrow 25r^2 = 288$ $\Rightarrow r^2 = \frac{288}{25}$.In $	riangle OPQ$,$PQ^2 = OP^2 + OQ^2 = r^2 + r^2 = 2r^2$.Then $l = OP^2 + PQ^2 + QO^2 = r^2 + 2r^2 + r^2 = 4r^2$.$l = 4 	imes \frac{288}{25} = \frac{1152}{25} = 46.08$.The greatest integer less than or equal to $l$ is $\lfloor 46.08 floor = 46$.

Let two straight lines drawn from the origin $O$ intersect the line $3x + 4y = 12$ at the points $P$ and $Q$ such that $\triangle OPQ$ is an isosceles triangle and $\angle POQ = 90^{\circ}$. If $l = OP^2 + PQ^2 + QO^2$,then the greatest integer less than or equal to $l$ is:

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