Find the ratio in which $2 x + 3 y + 5 z = 1$ divides the line joining the points $(1, 0, - 3)$ and $(1, - 5, 7)$ .

1 : 2

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2 : 1

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3 : 2

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2 : 3

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Solution

The correct option is D $2 : 3$

$2 x + 3 y + 5 z = 1$ divides $(1, 0, - 3)$ and $(1, - 5, 7)$ in the ratio of $k : 1$ at point $P$ .

Then, $P (\frac{k + 1}{k + 1}, \frac{- 5 k}{k + 1}, \frac{7 k - 3}{k + 1})$ which must satisfy $2 x + 3 y + 5 z = 1$

$\Rightarrow 2 (\frac{k + 1}{k + 1}) + 3 (\frac{- 5 k}{k + 1}) + 5 (\frac{7 k - 3}{k + 1}) = 1$

$\Rightarrow 2 k + 2 - 15 k + 35 k - 15 = k + 1$

$\Rightarrow 21 k = 14$

$\Rightarrow k = \frac{2}{3}$

$∴ 2 x + 3 y + 5 z = 1$ divides $(1, 0, - 3)$ and $(1, - 5, 7)$ in the ratio of $2 : 3$ .

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