Let $L_{1} : \frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0}$ and $L_{2} : \frac{x - 2}{2} = \frac{y}{0} = \frac{z + 4}{α}, α \in R,$ be two lines, which intersect at the point B. If P is the foot of perpendicular from the point A(1, 1, –1) on L₂, then the value of 26 α(PB)² is ____.

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answer is 216.

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Detailed Solution

Point B
$\begin{array}{l} (3 λ + 1, - λ + 1, - 1) \equiv (2 μ + 2, 0, αμ - 4) \\ 3 λ + 1 = 2 μ + 2 \\ - λ + 1 = 0 \\ - 1 = αμ - 4 \\ λ = 1, μ = 1, α = 3 \\ B (4, 0, - 1) \end{array}$
Let Point ‘P’ is $(2 δ + 2, 0, 3 δ - 4)$
Dr’s of AP $< 2 δ + 1, - 1, 3 δ - 3 >$
$\begin{array}{l} AP ⊥ L_{2} \Rightarrow δ = \frac{7}{13} \\ P (\frac{40}{13}, 0, \frac{- 31}{13}) \\ 2 σ δ (PB)^{2} = 26 \times 3 \times (\frac{144}{169} + \frac{324}{169}) \\ = 216 \end{array}$