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Q.
If the lines $\frac{x+1}{2} = \frac{y-1}{1} = \frac{z+1}{3} $ and $\frac{x+2}{2} = \frac{y-k}{3} =\frac{z}{4} $ are coplanar, then the value of k is :
Three Dimensional Geometry
Solution:
Two given planes are coplanar, if
$\begin{vmatrix}-2-\left(-1\right)&k-1&0-\left(-1\right)\\ 2&1&3\\ 2&3&4\end{vmatrix}= 0$
$ \Rightarrow \begin{vmatrix}-1&k-1&1\\ 2&1&3\\ 2&3&4\end{vmatrix} = 0$
$\Rightarrow \left(- 1\right) \left(4 - 9\right) - \left(k - 1\right) \left(8 - 6\right) + 6 - 2 = 0 $
$\Rightarrow k = \frac{11}{2} $