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The square of the distance of the image of the point $(6,1,5)$ in the line $\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4}$, from the origin is _________
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The correct answer is:
62

Let $\mathrm{M}(3 \lambda+1,2 \lambda, 4 \lambda+2)$
$\overrightarrow{\mathrm{AM}} \cdot \overrightarrow{\mathrm{b}}=0$
$\Rightarrow \quad 9 \lambda-15+4 \lambda-2+16 \lambda-12=0$
$\Rightarrow \quad 29 \lambda=29$
$\Rightarrow \quad \lambda=1$
$\mathrm{M}(4,2,6), \mathrm{I}=(2,3,7)$
Required Distance $=\sqrt{4+9+49}=\sqrt{62}$
Ans. 62
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