"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 :"

Question

Accepted Answer

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$$\begin{aligned} & P_1:(3 k-6,2 k, k-1) \\ & P_2(4 \alpha+7,3 \alpha+9,2 \alpha+4) \\ & P_1 \equiv P_2 \\ & 3 k-6=4 \alpha+7 \Rightarrow 3 k-4 \alpha=13 \\ & 2 k=3 \alpha+9 \Rightarrow 2 k-3 \alpha=9 \\ & \therefore k=3, \alpha=-1 \\ & \therefore P_1:(3,6,2) \end{aligned}$$

Distance of $(3,6,2)$ and $(7,8,9)$

$$\begin{aligned} & =\sqrt{16+4+49}=\sqrt{69}=d \\ & d^2+6=69+6=75 \end{aligned}$$

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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 :

Solution

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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 :

Solution

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