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If the straight lines $x=1+s, y=-3-\lambda s, z=1+\lambda s$ and $x=\frac{t}{2}, y=1+t, z=2-t$ with parameters $s$ and $t$ respectively, are co-planar then $\lambda$ equals
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$-2$
$-2$
Given lines $\frac{x-1}{1}=\frac{y+s}{-\lambda}=\frac{z-1}{\lambda}=s$ and $\frac{x}{1 / 2}=\frac{y-1}{1}=\frac{z-2}{-1}=t$ are coplanar then plan passing through these lines has normal perpendicular to these lines $\Rightarrow a-b \lambda+c \lambda=0 \quad$ and $\frac{a}{2}+b-c=0$ (where $a, b, c$ are direction ratios of the normal to the plan) On solving, we get $\lambda=-2$
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