The lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k} \quad$ and…

The lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k} \quad$ and $\frac{x-1}{\mathrm{k}}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar if
  1. $\mathrm{k}=1$ or $\mathrm{k}=-1$
  2. $\mathrm{k}=0$ or $\mathrm{k}=-3$
  3. $\mathrm{k}=3$ or $\mathrm{k}=-3$
  4. $\mathrm{k}=0$ or $\mathrm{k}=3$

Solution

If the given lines are co-planar, we get $\left|\begin{array}{ccc}2-1 & 3-4 & 4-5 \\ 1 & 1 & -k \\ k & 2 & 1\end{array}\right|=0$ $\begin{array}{ll}\therefore & \left|\begin{array}{ccc}1 & -1 & -1 \\ 1 & 1 & -k \\ k & 2 & 1\end{array}\right|=0 \\ \therefore & 1(1+2 k)+1\left(1+k^2\right)-1(2-k)=0 \\ \therefore & 1+2 k+1+k^2-2+k=0 \\ \therefore & k^2+3 k=0 \\ \therefore & k=0 \text { or } k=-3\end{array}$

Asked in: MHT CET 2024 (04 May Shift 1)

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