The distance of point of intersection of lines x-4 1 = y+3 -4 = z+1 7 and x-1 2 = y+1 -3…
Mathematics · JEE Main · NTA Exams — Three Dimensional Geometry
The distance of point of intersection of lines
\(\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7} \text { and } \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}\) from the point \((1,-4,7)\)
- 6
- \(\sqrt{24}\)
- \(\sqrt{26}\)
- none of these
Answer
(C) 26
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