The distance of the point (1, 1, 9) from the point of intersection of the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ and the plane $x + y + z = 17$ is :
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- $2\sqrt{19}$
- $19\sqrt{2}$
- $\sqrt{38}$
- 38
The Correct Option is C
Solution and Explanation
Step 2: Substitute in plane: $(r+3) + (2r+4) + (2r+5) = 17 \Rightarrow 5r = 5 \Rightarrow r = 1$.
Step 3: Intersection point: $(4, 6, 7)$.
Step 4: Distance from (1, 1, 9): $d = \sqrt{(4-1)^2 + (6-1)^2 + (7-9)^2} = \sqrt{9 + 25 + 4} = \sqrt{38}$.
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