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3D Geometry question

2024 · 30 Jan · Shift 2 · Q55
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  5. /2024 · 30 Jan · Shift 2 · Q55

3D Geometry question

2024 · 30 Jan · Shift 2 · Q55

JEE MainMathematics3D GeometryNumerical+4 / −1
Let a line passing through the point (−1,2,3)(-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}L1​:3x−1​=2y−2​=−2z+1​ at M(α,β,γ)M(\alpha, \beta, \gamma)M(α,β,γ) and L2:x+2−3=y−2−2=z−14L_2: \frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}L2​:−3x+2​=−2y−2​=4z−1​ at N(a,b,c)N(a, b, c)N(a,b,c). Then, the value of (α+β+γ)2(a+b+c)2\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}(a+b+c)2(α+β+γ)2​ equals ‾\underline{\hspace{2cm}}​.
Numerical answer
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