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In $\triangle ABC$, the midpoints of the sides $AB, BC$, and $CA$ are respectively $(l, 0, 0), (0, m, 0)$, and $(0, 0, n)$. Then, $\frac{AB^2+BC^2+CA^2}{l^2+m^2+n^2}$ is equal to

Suppose $A$ and $B$ are the points at which the line $x+y-\lambda=0$ meets the pair of straight lines $x^2+y^2-2x-4y+2=0$. If $\angle AOB=90^{\circ}$,then a value of $\lambda$ is

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