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(D) Let $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} = \lambda$. Any point $P$ on the line is given by $(3\lambda+6, 2\lambda+7, -2\lambda+7)$. Let

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$7$

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(D) Let $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} = \lambda$. Any point $P$ on the line is given by $(3\lambda+6, 2\lambda+7, -2\lambda+7)$. Let $A = (1, 2, 3)$. The direction ratios of the line $AP$ are $(3\lambda+6-1, 2\lambda+7-2, -2\lambda+7-3)$, which simplifies to $(3\lambda+5, 2\lambda+5, -2\lambda+4)$. Since $AP$ is perpendicular to the given line with direction ratios $(3, 2, -2)$, the dot product of their direction ratios must be zero: $3(3\lambda+5) + 2(2\lambda+5) - 2(-2\lambda+4) = 0$. $9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0$. $17\lambda + 17 = 0 \Rightarrow \lambda = -1$. Substituting $\lambda = -1$ into the coordinates of $P$, we get $P = (3( -1)+6, 2(-1)+7, -2(-1)+7) = (3, 5, 9)$. The length of the perpendicular $AP$ is the distance between $A(1, 2, 3)$ and $P(3, 5, 9)$: $AP = \sqrt{(3-1)^2 + (5-2)^2 + (9-3)^2} = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 	ext{ units}$.

The length of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$ is (in $\text{ units}$)

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