Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The condition for coplanarity of two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ is given by:

$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$

Substituting the values from the given lines:

$\left|\begin{array}{ccc}1-2& 4-3& 5-4\\ 1& 1& -k\\ k& 2& 1\end{array}\right|=0$

$\left|\begin{array}{ccc}-1& 1& 1\\ 1& 1& -k\\ k& 2& 1\end{array}\right|=0$

Expanding the determinant along the first row:

$-1(1-(-2k))-1(1-(-k^2))+1(2-k)=0$

$-1(1+2k)-1(1+k^2)+1(2-k)=0$

$-1-2k-1-k^2+2-k=0$

$-k^2-3k=0$

$k(k+3)=0$

$\Rightarrow k=0$ or $k=-3$

Answer: $\mathrm{k}=0$ or $\mathrm{k}=-3$