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If the lines $\frac{x - 1}{2} = \frac{y - 2}{x_{1}} = \frac{z - 3}{x_{2}}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ lie in the same plane, then for the equation $x_{1} t^{2} + (x_{2} + 2) t + a = 0,$ sum of roots of equation is

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-1

-2

answer is D.

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Detailed Solution

$[\begin{matrix} a - c & b & d \end{matrix}] = 0 \Rightarrow |\begin{array}{ccc} 1 & 1 & 1 \\ 2 & x_{1} & x_{2} \\ 3 & 4 & 5 \end{array}| = 0$

$\Rightarrow 5 x_{1} - 4 x_{2} - (10 - 3 x_{2}) + (8 - 3 x_{1}) = 0$

$\Rightarrow 2 x_{1} - x_{2} - 2 = 0 \Rightarrow x_{2} + 2 = 2 x_{1}$

$∴ sum of roots = - 2$

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