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A line through a given point meeting two given lines

This type of question, asked 3 times: 30 Jan 2024, 3 Apr 2025, 7 Apr 2025. All 3 questions below, oldest first.

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  1. Question 1

    30 Jan 2024

    172
    Let a line passing through the point (−1,2,3)( - 1,2,3) intersect the lines L1:x−13=y−22=z+1−2L_{1}:\frac{x - 1}{3}=\frac{y - 2}{2}=\frac{z + 1}{- 2} at M(α,β,γ)M(\alpha,\beta,\gamma) and L2:x+2−3=y−2−2=z−14L_{2}:\frac{x + 2}{- 3}=\frac{y - 2}{- 2}=\frac{z - 1}{4} at N(a,bN(a,b,c). Then the value of (α+β+γ)2(a+b+c)2\frac{(\alpha + \beta + \gamma)^{2}}{(a + b + c)^{2}} equals

  2. Question 2

    3 Apr 2025

    52
    Let a line passing through the point (4,1,0)(4,1,0) intersect the line L1;x−12=y−23=z−34L_{1};\frac{x- 1}{2}=\frac{y- 2}{3}=\frac{z- 3}{4} at the point A (α,β,γ)(\alpha,\beta,\gamma) and the line L2:x−6=y=−z+4L_{2}:x- 6 =y= -z+ 4 at the point B(a,b,c)B(a,b,c). Then ∣101αβγabc∣\left| \begin{matrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{matrix} \right| is equal to

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  3. Question 3

    7 Apr 2025

    68
    Let the line L pass through (1,1,1)(1,1,1) and intersect the lines x−12=y+13=z−14\frac{x- 1}{2}=\frac{y+ 1}{3}=\frac{z- 1}{4} and x−31=y−42=z1\frac{x- 3}{1}=\frac{y- 4}{2}=\frac{z}{1} Then, which of the following points lies on the line L?.

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