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JEE Main202624 January 2026Evening ShiftMathematicsThree Dimensional GeometryActual

The sum of all values of , for which the shortest distance between the lines x+1 = y-2 -1 = z-4 - and x = y-1 2 = z-1 2 is 2 , is

Options

  1. A6
  2. B8
  3. C-8
  4. D-6

Correct answer

D. -6

Step-by-step solution

The given lines are L₁: x+1 = y-2 -1 = z-4 - and L₂: x = y-1 2 = z-1 2 . For L₁ , point a ₁ = (-1, 2, 4) and direction b ₁ = ( , -1, - ) . For L₂ , point a ₂ = (0, 1, 1) and direction b ₂ = ( , 2, 2 ) . The shortest distance d is given by d = |( a ₂ - a ₁) ( b ₁ b ₂)| | b ₁ b ₂| . a ₂ - a ₁ = (1, -1, -3) . b ₁ b ₂ = vmatrix i & j & k & -1 & - & 2 & 2 vmatrix = i (-2 + 2 ) - j (2 ^2 + ^2) + k (2 + ) = (0, -3 ^2, 3 ) . | b ₁ b ₂| = 0 + 9 ^4 + 9 ^2 = 3| | ^2 + 1 . ( a ₂ - a ₁) ( b ₁ b ₂) = (1)(0) + (-1)(-3 ^2) + (-3)(

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