The distance of the point (1, 6, 2) from the | vedclass

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(C) Let $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} = \lambda$. Therefore, the coordinates of any point $P$ on the line are given by $P = (3\lambd

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$13$

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(C) Let $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} = \lambda$. Therefore, the coordinates of any point $P$ on the line are given by $P = (3\lambda + 2, 4\lambda - 1, 12\lambda + 2)$. Since this point $P$ lies on the plane $x - y + z = 16$, we substitute the coordinates into the plane equation: $(3\lambda + 2) - (4\lambda - 1) + (12\lambda + 2) = 16$. $3\lambda + 2 - 4\lambda + 1 + 12\lambda + 2 = 16$. $11\lambda + 5 = 16$. $11\lambda = 11$, which gives $\lambda = 1$. Substituting $\lambda = 1$ back into the coordinates of $P$, we get $P = (3(1) + 2, 4(1) - 1, 12(1) + 2) = (5, 3, 14)$. Now, we find the distance between $P(5, 3, 14)$ and $Q(1, 6, 2)$ using the distance formula: $PQ = \sqrt{(1-5)^2 + (6-3)^2 + (2-14)^2}$. $PQ = \sqrt{(-4)^2 + (3)^2 + (-12)^2}$. $PQ = \sqrt{16 + 9 + 144} = \sqrt{169} = 13 	ext{ units}$.

The distance of the point $(1, 6, 2)$ from the point of intersection of the line $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$ and the plane $x-y+z=16$ is (in $\text{ units}$)

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