The lines x−$$1a=y$$−$$23=z$$−$$34$$; x−$$23=y$$−$$34=z$$−$$45$$ are coplanar. Then a $$=$$

$$x2$$−$$x1y2$$−$$y1z2$$−$$z1l1m1n1l2m2n2=0$$

Straight lines in 3D

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Question

The lines $\frac{x - 1}{a} = \frac{y - 2}{3} = \frac{z - 3}{4}$ ; $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ are coplanar. Then a =

Moderate

Solution

$|\begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{matrix}|$ =0

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