If the line frac{x - 1}{2} = frac{y + 1}{3} = | vedclass

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(C) Any point on the given line can be represented as $(1 + 2\lambda, -1 + 3\lambda, 2 + 4\lambda)$ for some $\lambda \in \mathbb{R}$.Substituting thi

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$\frac{9}{2}$

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(C) Any point on the given line can be represented as $(1 + 2\lambda, -1 + 3\lambda, 2 + 4\lambda)$ for some $\lambda \in \mathbb{R}$.Substituting this point into the plane equation $x + 2y + 3z = 15$:$(1 + 2\lambda) + 2(-1 + 3\lambda) + 3(2 + 4\lambda) = 15$$1 + 2\lambda - 2 + 6\lambda + 6 + 12\lambda = 15$$20\lambda + 5 = 15$$20\lambda = 10$$\lambda = \frac{1}{2}$Substituting $\lambda = \frac{1}{2}$ back into the point coordinates:$P = (1 + 2(\frac{1}{2}), -1 + 3(\frac{1}{2}), 2 + 4(\frac{1}{2})) = (2, \frac{1}{2}, 4)$.The distance of $P(2, \frac{1}{2}, 4)$ from the origin $(0, 0, 0)$ is $\sqrt{2^2 + (\frac{1}{2})^2 + 4^2}$.$= \sqrt{4 + \frac{1}{4} + 16} = \sqrt{20 + \frac{1}{4}} = \sqrt{\frac{81}{4}} = \frac{9}{2}$.

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

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