The distance of the plane passing through (1,1,1) and perpendicular to the line $\frac{x - 1}{3} = \frac{y - 1}{0} = \frac{z - 1}{4}$ from the origin is

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Solution

The correct option is C

Equation of the plane is $3 (x - 1) + 0 (y - 1) + 4 (z - 1) = 0$

$\Rightarrow 3 x + 4 z - 7 = 0$

$∴ D i s t . f r o m o r i g i n = \frac{| - 7 |}{\sqrt{3^{2} + 4^{2}}} = \frac{7}{5}$

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The distance of the plane passing through (1,1,1) and perpendicular to the line x−13=y−10=z−14 from the origin is

The correct option is C Equation of the plane is 3(x−1)+0(y−1)+4(z−1)=0 ⇒ 3x+4z−7=0 ∴ Dist. from origin=|−7|√32+42=75

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The correct option is C

Equation of the plane is $3 (x - 1) + 0 (y - 1) + 4 (z - 1) = 0$

$\Rightarrow 3 x + 4 z - 7 = 0$

$∴ D i s t . f r o m o r i g i n = \frac{| - 7 |}{\sqrt{3^{2} + 4^{2}}} = \frac{7}{5}$