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If the line, $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane, $x + 2 y + 3 z = 15$ at a point P, then the distance of P from the origin is

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9/2

7/2

$2 \sqrt{5}$

$\frac{\sqrt{5}}{2}$

answer is A.

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Detailed Solution

$L e t \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4} = λ, P = (2 λ + 1, 3 λ - 1, 4 λ + 2)$ lines on the plane

$\Rightarrow λ = 1 / 2$

$∴ P = (2, \frac{1}{2}, 4), O P = \frac{9}{2}$

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