If the lines x−12=2−y3=z−3α and x−45=y−12=zβ intersect, then the magnitude of the minimum value of 8αβ is_______.

The lines x−12=y−23=z−3α and x−45=y−12=zβ are intersectx2−x1y2−y1z2−z1a1b1c1a2b2c2=04−11−20−323α52β=03−1−323α52β=03(3β−2α)+1(2β−5α)−3(4−15)=09β−6α+2β−5α+33=011β−11α+33=0α−β=3⇒α=3+β8αβ=8(3+β)β=83β+β2 Let f(β)=83β+β2 f′(β)=8(3+2β)f′′(β)=16f′(β)=0⇒β=-32f′′(β)β=−32=16>0f (β) is minimum of β=−32⇒α=3−32=32|8αβ|=832−32=|−18|=18

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If the lines $\frac{x - 1}{2} = \frac{2 - y}{3} = \frac{z - 3}{α} and \frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β}$ intersect, then the magnitude of the minimum value of 8 $αβ$ is_______.

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answer is 18.

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Detailed Solution

The lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{α} and \frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β}$ are intersect
$\begin{array}{l} |\begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}| = 0 \\ |\begin{array}{ccc} 4 - 1 & 1 - 2 & 0 - 3 \\ 2 & 3 & α \\ 5 & 2 & β \end{array}| = 0 \\ |\begin{array}{ccl} 3 & - 1 & - 3 \\ 2 & 3 & α \\ 5 & 2 & β \end{array}| = 0 \end{array}$
$\begin{array}{l} 3 (3 β - 2 α) + 1 (2 β - 5 α) - 3 (4 - 15) = 0 \\ 9 β - 6 α + 2 β - 5 α + 33 = 0 \\ 11 β - 11 α + 33 = 0 \\ α - β = 3 \\ \Rightarrow α = 3 + β \\ 8 αβ = 8 (3 + β) β \end{array}$
$= 8 (3 β + β^{2}) Let f (β) = 8 (3 β + β^{2}) \begin{array}{l} f^{'} (β) = 8 (3 + 2 β) \\ f^{''} (β) = 16 \\ f^{'} (β) = 0 \Rightarrow β = - \frac{3}{2} \\ {[f^{''} (β)]}_{β = - \frac{3}{2}} = 16 > 0 \end{array}$
f ( $β$ ) is minimum of $β = - \frac{3}{2}$
$\begin{array}{l} \Rightarrow α = 3 - \frac{3}{2} = \frac{3}{2} \\ | 8 αβ | = |8 (\frac{3}{2}) (- \frac{3}{2})| \\ = | - 18 | = 18 \end{array}$