If the lines, x-1 2 = y+1 3 = z-2 4 meets the plane, x+2 y+3 z=15 at a point P, then the…
Mathematics · JEE Main · NTA Exams — Three Dimensional Geometry
If the lines, \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}\) meets the plane, \(x+2 y+3 z=15\) at a point P, then the distance of P from the origin is:
- \(\sqrt{5} / 2\)
- \(\frac{7}{2}\)
- \(2 \sqrt{5}\)
- \(\frac{9}{2}\)
Answer
(B) 7 2
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