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Mathematics Question on 3D Geometry

Let a line passing through the point (−1,2,3)(-1, 2, 3) intersect the linesL1: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) andL2:x+2−3=y−2−2=z−14L_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 ____.

Answer

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 L1L_1: For intersection, equate:

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

Intersection with L2L_2: For intersection, equate:

−1+λl=−2−3ν,2+λm=2+4ν,3+λn=1−2ν.-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 cc.

Calculate:

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