MCQMediumJEE Main 2025 · 28 January, Shift 2Equation of Line in 3D

Mathematics Question from JEE Main 2025 · 28 January, Shift 2

The square of the distance of the point (157,327,7)\left( \frac{15}{7}, \frac{32}{7}, 7 \right) from the line x+13=y+35=z+57\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7} in the direction of the vector i+4j+7k\mathbf{i} + 4\mathbf{j} + 7\mathbf{k} is:

  • A

    4141

  • B

    4444

  • C

    5454

  • D

    6666

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