MCQMediumJEE Main 2025 · 4 April, Shift 2Cross Product

Mathematics Question from JEE Main 2025 · 4 April, Shift 2

Let A be the point of intersection of the lines L1:x71=y50=z31andL2:x13=y+34=z+75L_1 : \frac{x - 7}{1} = \frac{y - 5}{0} = \frac{z - 3}{-1} \quad \text{and} \quad L_2 : \frac{x - 1}{3} = \frac{y + 3}{4} = \frac{z + 7}{5} Let B and C be the points on the lines L1L_1 and L2L_2, respectively, such that AB=AC=15AB = AC = \sqrt{15}. Then the square of the area of the triangle ABC is:

  • A

    5454

  • B

    6363

  • C

    5757

  • D

    6060

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