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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Aicharya · Accepted Answer

$\begin{aligned} & \frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}=\lambda....(1)\ & x=3 \lambda-6, y=2 \lambda, z=\lambda-1 \ & \frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}=\mu....(2)\end{aligned}$ $\mathrm{x}=4 \mu+7, \mathrm{y}=3 \mu+9, \mathrm{z}=2 \mu+4$ $3 \lambda-6=4 \mu+7 \Rightarrow 3 \lambda-4 \mu=13$ $\ldots(3) 	imes 2$ $2 \lambda=3 \mu+9 \Rightarrow 2 \lambda-3 \mu=9$ $\ldots(4) 	imes 3$ $\begin{array}{c} 6 \lambda-8 \mu=26 \ 6 \lambda-9 \mu=27 \ -\quad+\quad- \ \hline\mu=-1 \end{array}$ $\begin{gathered} \Rightarrow 3 \lambda-4(-1)=13 \ 3 \lambda=9 \ \lambda=3 \end{gathered}$ int. point $(3,6,2) ;(7,8,9)$ $\mathrm{d}^2=16+4+49=69$ Ans. $\mathrm{d}^2+6=69+6=75$

Let $\mathrm{d}$ be the distance of the point of intersection of the lines…

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