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+1 vote
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in Mathematics by (50.3k points)
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Let 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 :

(1) 72

(2) 69

(3) 75

(4) 78

1 Answer

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Best answer

Correct option is (3) 75

\(\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}=\lambda \quad ....(1)\)

\(\mathrm{x}=3 \lambda-6, \mathrm{y}=2 \lambda, \mathrm{z}=\lambda-1\)

\(\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}=\mu \quad....(2)\)

\(\mathrm{x}=4 \mu+7, \mathrm{y}=3 \mu+9, \mathrm{z}=2 \mu+4\)

\(3 \lambda-6=4 \mu+7 \Rightarrow 3 \lambda-4 \mu=13 \quad....(3) \times 2\)

\(2 \lambda=3 \mu+9 \Rightarrow 2 \lambda-3 \mu=9 \quad....(4) \times 3\)

\(\begin{array} {l}6 \lambda-8 \mu=26\\ 6 \lambda-9 \mu=27\\ -\quad+\quad-\\\hline\quad\mu=-1 \end{array}\)

\(\Rightarrow 3 \lambda-4(-1)=13\)

\(3 \lambda=9\)

\(\lambda=3\)

int. point \((3,6,2) ;(7,8,9)\)

\(\mathrm{d}^{2}=16+4+49=69\)

\(\mathrm{d}^{2}+6=69+6=75\)

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