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Unit vector perpendicular to L1: (x+1)/3 = (y+2)/1 = (z+1)/2 and L2: (x-2)/1 = (y+2)/2 = (z-3)/3

The same question, asked 3 times: 3 May 2023 S2*, 4 May 2023 S1, 14 May 2024 S1. Every wording below, oldest first.

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  1. Question 1

    3 May 2023 S2

    117
    Let L1:x+13=y+22=z+11L_1:\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+1}{1} and L2:x−22=y+21=z−33L_2:\frac{x-2}{2}=\frac{y+2}{1}=\frac{z-3}{3} be the given lines. Then the unit vector perpendicular to L1L_1 and L2L_2 is

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  2. Question 2

    4 May 2023 S1

    129
    Let L1:x+13=y+21=z+12L_1:\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2} and L2:x−21=y+22=z−33L_2:\frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3} be two given lines. Then the unit vector perpendicular to L1L_1 and L2L_2 is

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  3. Question 3

    14 May 2024 S1

    116
    Consider the lines L1:x+13=y+21=z+12L_1:\dfrac{x+1}{3}=\dfrac{y+2}{1}=\dfrac{z+1}{2} and L2:x−21=y+22=z−33L_2:\dfrac{x-2}{1}=\dfrac{y+2}{2}=\dfrac{z-3}{3}, then the unit vector perpendicular to both L1L_1 and L2L_2 is

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