Let \mathrm{d} be the distance of the point of intersection of the lines \frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1} and \frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2} from the point (7,8,9). Then \mathrm{d}^2+6 is equal to :

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Answer: A. 75 \begin{aligned} & P_1:(3 k-6,2 k, k-1) \ & P_2(4 \alpha+7,3 \alpha+9,2 \alpha+4) \ & P_1 \equiv P_2 \ & 3 k-6=4 \alpha+7 \Rightarrow 3 k-4 \alpha=13 \ & 2 k=3 \alpha+9 \Rightarrow 2 k-3 \alpha=9 \ & 	herefore k=3, \alpha=-1 \ & 	herefore P_1:(3,6,2) \end{aligned} Distance of (3,6,2) and (7,8,9) \begin{aligned} & =\sqrt{16+4+49}=\sqrt{69}=d \ & d^2+6=69+6=75 \end{aligned}

Let $$\mathrm{d}$$ be the distance of the point of intersection of the lines $$\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$$ and $$\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$$ from the point $$(7,8,9)$$. Then $$\mathrm{d}^2+6$$ is equal to :

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