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 $k$ is______.

Answer 1

Answer: 1

$\left|\begin{array}{ccc}k+1& 4& 6\\ 1& 2& 3\\ 3& 2& 1\end{array}\right|=0$
$(k+1)[2-6]-4[1-9]+6[2-6]=0$
$k=1$