Question:
If the lines $\frac{x-k}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and
$\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+3}{1}$ are co-planar, then the value
of $\mathrm{k}$ is________.
Solution:
$\left|\begin{array}{ccc}\mathrm{k}+1 & 4 & 6 \\ 1 & 2 & 3 \\ 3 & 2 & 1\end{array}\right|=0$
$(k+1)[2-6]-4[1-9]+6[2-6]=0$
$\mathrm{k}=1$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.