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

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a

34

b

43

c

75

d

1

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detailed solution

Correct option is C

Equation of the plane is a(x−1)+b(y−1)+c(z−1)=0…….(1)Since the line is perpendicular to the plane (1) ∴3(x−1)+0(y−1)+4(z−1)=0⇒3x+0y+4z−7=0……(2)Distance from origin (0, 0, 0) to equation (2) is d=ax1+by1+cz1+da2+b2+c2∴d=−75=75 Therefore, the correct answer is (3).

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The equation of a plane which passes through the point of intersection of lines $\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ and at greatest distance from point (0,0,0) is

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