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Point of intersection of two lines, then a distance

This type of question, asked 6 times: 27 Jan 2024, 29 Jan 2024, 4 Apr 2024, 5 Apr 2024, 24 Jan 2025, 5 Apr 2026 S1. All 6 questions below, oldest first.

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  1. Question 1

    27 Jan 2024

    179
    The lines x−22=y−2=z−716\frac{x- 2}{2}=\frac{y}{- 2}=\frac{z- 7}{16} and x+34=y+23=z+21\frac{x+ 3}{4}=\frac{y+ 2}{3}=\frac{z+ 2}{1} intersect at the point PP. If the distance of PP from the line x+12=y−13=z−11\frac{x + 1}{2}=\frac{y - 1}{3}=\frac{z - 1}{1} is ll, then 14l214l^{2} is equal to.......

  2. Question 2

    29 Jan 2024

    76
    Let PQRPQR be a triangle with R(−1,4,2)R( - 1,4,2). Suppose M(2,1,2)M(2,1,2) is the mid point of PQPQ. The distance of the centroid of △PQR\triangle PQR from the point of intersection of the lines x−20=y2=z+3−1\frac{x - 2}{0}=\frac{y}{2}=\frac{z + 3}{- 1} and x−11=y+3−3=z+11\frac{x - 1}{1}=\frac{y + 3}{- 3}=\frac{z + 1}{1} is

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  3. Question 3

    4 Apr 2024

    170
    Let PP be the point of intersection of the lines x−21=y−45=z−21\frac{x - 2}{1}=\frac{y - 4}{5}=\frac{z - 2}{1} and  x−32=y−23=z−32\ \frac{x - 3}{2}=\frac{y - 2}{3}=\frac{z - 3}{2}. Then, the shortest distance of PP from the line 4x=2y=z4x = 2y = z is

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  4. Question 4

    5 Apr 2024

    61
    Let dd be the distance of the point of intersection of the lines x+63=y2=z+11\frac{x + 6}{3}=\frac{y}{2}=\frac{z + 1}{1} and x−74=y−93=z−42\frac{x- 7}{4}=\frac{y- 9}{3}=\frac{z- 4}{2} from the point (7,8,9)(7,8,9). Then d2+6d^{2}+ 6 is equal to:

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  5. Question 5

    24 Jan 2025

    68
    Let the line passing through the points (−1,2,1)( - 1,2,1) and parallel to the line x−12=y+13=z4\frac{x - 1}{2}=\frac{y + 1}{3}=\frac{z}{4} intersect the line x+23=y−32=z−41\frac{x + 2}{3}=\frac{y - 3}{2}=\frac{z - 4}{1} at the point PP. Then the distance of P from the point Q(4,−5,1)Q(4, - 5,1) is :

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  6. Question 6

    5 Apr 2026 S1

    65
    The square of the distance of the point of intersection of the line r→=(i^+j^−k^)+λ(a^−j^)\overrightarrow{r}= (\widehat{i}+\widehat{j}-\widehat{k}) +\lambda(\widehat{a}-\widehat{j}), a≠0a \neq 0 and r→=(4i^−k^)+μ(2i^+ak^)\overrightarrow{r}= (4\widehat{i}-\widehat{k}) +\mu(2\widehat{i}+ a\widehat{k}) from the origin is :

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