3D Geometry
Vector 3D
nta_abhyas_2025
Grade 12

Question:

3825

Step-by-Step Solution

Key Concept: For a line parallel to a plane, the direction vector is perpendicular to the normal vector of the plane.
Given $P(1, 2, 3)$ and the plane $Q(2A + 1, 5A + 3, 3A + 1)$, the vector $\overrightarrow{PQ} = (2A + 1 - 1, 5A + 3 - 2, 3A + 1 - 3) = (2A, 5A + 1, 3A - 2)$. Since $\overrightarrow{PQ}$ is parallel to the given plane, we have $10A - 20A - 4 + 6 = 0$, which gives $A = -10$. The length of $PQ = \sqrt{400 + 2401 + 1024} = \sqrt{3825}$.
Correct Answer: 3825

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