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The triangle ABC is such that the midpoints of the sides BC,CA, AB are $(l, 0, 0), (0, m, 0) (0, 0, n) r e s p e c t i v e l y t h e n$ $\frac{A B^{2} + B C^{2} + C A^{2}}{l^{2} + m^{2} + n^{2}} =$

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Detailed Solution

$\frac{{AB}^{2} + {BC}^{2} + {CA}^{2}}{l^{2} + m^{2} + n^{2}} = \frac{4 ({DE}^{2} + {EF}^{2} + {DF}^{2})}{l^{2} + m^{2} + n^{2}}$

$= \frac{4.2 (l^{2} + m^{2} + n^{2})}{(l^{2} + m^{2} + n^{2})} = 8$

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