Question

If the lines $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-k}{5}$ and $\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$ are coplanar, then the value of $k$ is

Answer 1

Points on the given two lines can be taken as $(x_1,y_1,z_1)=(-3,1,k)$ and $(x_2,y_2,z_2)=(-1,2,5)$
The corresponding $D.R^{\prime }s$ are $(a_1,b_1,c_1)=(-3,1,5)$ and $(a_2,b_2,c_2)=(-1,2,5)$
For the lines to be coplanar, $D=\left|\begin{array}{ccc}{x}_1-x_2& y_1-y_2& z_1-z_2\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}-2& -1& k-5\\ -3& 1& 5\\ -1& 2& 5\end{array}\right|=0$
$\Rightarrow -2(5-10)+1(-15+5)+(k-5)(-6+1)=0$
$\Rightarrow 10-10-5(k-5)=0$
$\Rightarrow k=5$