# The distance of the point P(3, 8, 2) from the line (x-1)/2=(y-3)/4=(z-2)/3, measured parallel to the plane 3x+2y-2z+15=0 is .

## Question

The distance of the point $P(3, 8, 2)$ from the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$, measured parallel to the plane $3x+2y-2z+15=0$ is $\quad$.

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## Answer

To find the **distance of the point ( P(3, 8, 2) ) from the line** $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$, measured parallel to the plane $ 3x+2y-2z+15=0 $, we follow these steps:

**Parameterize the Line**: Given the line equation $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3} = \lambda$, we substitute $\lambda$ to find a general point $ A $ on the line: $ x = 2\lambda + 1, \quad y = 4\lambda + 3, \quad z = 3\lambda + 2 $ So, the point $ A $ on the line is $ A(2\lambda + 1, 4\lambda + 3, 3\lambda + 2) $**Determine Correct (\lambda) for Measurement Conformity**: The distance from $ A $ to the given plane must equal the distance from point $ P $ to the plane, so we solve for $ \lambda $: $ \frac{|3(2\lambda + 1) + 2(4\lambda + 3 - 2(3\lambda + 2) + 15|}{\sqrt{3^2 + 2^2 + (-2)^2}} = \frac{|3(3) + 2(8) - 2(2) + 15|}{\sqrt{3^2 + 2^2 + (-2)^2}} $ Simplifying both sides, after calculation, it is found that $\lambda = 2$.**Calculate Distance ( PA )**: Substitute $\lambda = 2$ into $ A $ to get: $$ A = (2 \cdot 2 + 1, 4 \cdot 2 + 3, 3 \cdot 2 + 2) = (5, 11, 8) $$ The distance formula for points ( P(3, 8, 2) ) and ( A(5, 11, 8) ) is: $$ PA = \sqrt{(5 - 3)^2 + (11 - 8)^2 + (8 - 2)^2} = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{49} = 7 $$

Thus, the **distance** from $ P(3, 8, 2) $ to the line, measured parallel to the plane, is **7 units**.

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