Let a line passing through the point (−1,2,3) intersect the lines L1:3x−1=2y−2=−2z+1 at M(α,β,γ) and L2:−3x+2=−2y−2=4z−1 at N(a,b,c). Then the value of (a+b+c)2(α+β+γ)2 equals:
Answer: 196
Solutions · 1
Official solution
M=(3t+1,2t+2,−2t−1) on L1; N=(−3s−2,−2s+2,4s+1) on L2. Collinearity with P=(−1,2,3) (PM=kPN) gives k=−2,t=2s, and substituting into the z-equation: s=2,t=4.
\[M=(13,10,-9),\qquad N=(-8,-2,9).\]
\[\alpha+\beta+\gamma=13+10-9=14,\qquad a+b+c=-8-2+9=-1.\]
\[\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}=\frac{196}{1}=196.\]