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JEE Mains · Maths · STD 12 - 11. three dimension geometry

A line passes through \(A(4,-6,-2)\) and \(B(16,-2,4)\). The point \(\mathrm{P}(\mathrm{a}, \mathrm{b}, \mathrm{c})\) where \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are non-negative integers, on the line \(\mathrm{AB}\) lies at a distance of 21 units, from the point \(\mathrm{A}\). The distance between the points \(\mathrm{P}(\mathrm{a}, \mathrm{b}, \mathrm{c})\) and \(\mathrm{Q}(4,-12,3)\) is equal to ...........

  1. A \(19\)
  2. B \(21\)
  3. C \(20\)
  4. D \(22\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(22\)

Step-by-step Solution

Detailed explanation

\(\frac{x-4}{12}=\frac{x+6}{4}=\frac{z+2}{6} \) \(\frac{x-4}{\frac{6}{7}}=\frac{y+6}{\frac{2}{7}}=\frac{z+2}{\frac{3}{7}}=21 \) \(\left(21 \times \frac{6}{7}+4, \frac{2}{7} \times 21-6, \frac{3}{7} \times 21-2\right) \) \(=(22,0,7)=(a, b, c) \) \(\therefore \sqrt{324+144+16}=22\)