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

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

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

Substituting the given values $(x_1,y_1,z_1)=(3,2,5)$, $(x_2,y_2,z_2)=(4,3,3)$, $(a_1,b_1,c_1)=(1,1,-k)$, and $(a_2,b_2,c_2)=(k,1,2)$:

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

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

Expanding the determinant along the first row:

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

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

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

$-k^2+3k-2=0$

$k^2-3k+2=0$

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

$k=1,2$

Answer: 1,2