If a plane passes through the point (1, 1, 1) and is perpendicular to the line x−13=y−10=z−14 then its perpendicular distance from the origin is

Equation of plane through (1, 1, 1) and perpendicular to the linex−13=y−10=z−14 is 3x+4z−7=0Distance of plane from the origin=0+0−79+16=75

If a plane passes through the point (1, 1, 1) and is perpendicular to the line x−13=y−10=z−14 then its perpendicular distance from the origin is

Equation of plane through (1, 1, 1) and perpendicular to the linex−13=y−10=z−14 is 3x+4z−7=0Distance of plane from the origin=0+0−79+16=75

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If a plane passes through the point (1, 1, 1) and is perpendicular to the line $\frac{x - 1}{3} = \frac{y - 1}{0} = \frac{z - 1}{4}$ then its perpendicular distance from the origin is

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$\frac{3}{4}$

$\frac{7}{5}$

$\frac{4}{3}$

$1$

answer is C.

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Detailed Solution

Equation of plane through (1, 1, 1) and perpendicular to the line

$\frac{x - 1}{3} = \frac{y - 1}{0} = \frac{z - 1}{4}$ is $3 x + 4 z - 7 = 0$

Distance of plane from the origin

$= |\frac{0 + 0 - 7}{\sqrt{9 + 16}}| = \frac{7}{5}$

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If a plane passes through the point (1, 1, 1) and is perpendicular to the line x−13=y−10=z−14 then its perpendicular distance from the origin is

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If a plane passes through the point (1, 1, 1) and is perpendicular to the line $\frac{x - 1}{3} = \frac{y - 1}{0} = \frac{z - 1}{4}$ then its perpendicular distance from the origin is