Question:

Let a line passing through the point (1,2,3)(-1, 2, 3) intersect the linesL1:x13=y22=z+12 L_1 : \frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z + 1}{-2} at M(α,β,γ) M(\alpha, \beta, \gamma) andL2:x+23=y22=z14 L_2 : \frac{x + 2}{-3} = \frac{y - 2}{-2} = \frac{z - 1}{4} at 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} equals ____.

Updated On: Mar 20, 2025
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Correct Answer: 196

Solution and Explanation

Write the equation of the line passing through (1,2,3) (-1, 2, 3) with direction ratios (l,m,n) (l, m, n) :

x=1+λl,y=2+λm,z=3+λn. x = -1 + \lambda l, \quad y = 2 + \lambda m, \quad z = 3 + \lambda n.

Intersection with L1 L_1 : For intersection, equate:

1+λl=1+3μ,2+λm=2+2μ,3+λn=12μ. -1 + \lambda l = 1 + 3\mu, \quad 2 + \lambda m = 2 + 2\mu, \quad 3 + \lambda n = -1 - 2\mu.

Intersection with L2 L_2 : For intersection, equate:

1+λl=23ν,2+λm=2+4ν,3+λn=12ν. -1 + \lambda l = -2 - 3\nu, \quad 2 + \lambda m = 2 + 4\nu, \quad 3 + \lambda n = 1 - 2\nu.

Solve for α,β,γ,a,b, \alpha, \beta, \gamma, a, b, and c c .

Calculate:

(α+β+γ)2a+b+c=196. \frac{(\alpha + \beta + \gamma)^2}{a + b + c} = 196.

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