Equation of plane parallel to $3 x + 4 y + 5 z - 6 = 0$ , $6 x + 8 y + 10 z - 16 = 0$ and equidistant from them is

3 x + 4 y + 5 z = 7

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3 x + 4 y + 5 z = 10

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6 x + 8 y + 10 z = 0

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6 x + 8 y + 10 z = 3

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Solution

The correct option is C $3 x + 4 y + 5 z = 7$
$P L_{1} : 3 x + 4 y + 5 z - 6 = 0 P L_{2} : 3 x + 4 y + 5 z - 8 = 0$
distance between $P L_{1}$ and $P L_{2}$ is $\frac{2}{\sqrt{3^{2} + 4^{2} + 5^{2}}} = \frac{2}{\sqrt{50}}$
Equidistant plane will be between two plane $P L_{1}$ and $P L_{2}$ at a distance $\frac{1}{\sqrt{50}}$ from both plane.
$∴ 3 x + 4 y + 5 z = k$ be the equation of plane.
$\frac{k - 6}{\sqrt{50}} = \frac{1}{\sqrt{50}} \Rightarrow k = 7 ∴ 3 x + 4 y + 5 z = 7$

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