Q.If lines $\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-1}{4}$ and $\dfrac{x-2}{1} = \dfrac{y+m}{2} = \dfrac{z-2}{1}$ intersect each other then find $m$.

Two lines intersect each other if and only if the shortest distance between them is zero. Applying Theorem 6.6's formula, lines $\vec r=\vec a_1+\lambda_1\vec b_1$ and $\vec r=\vec a_2+\lambda_2\vec b_2$ intersect if and only if

i.e. the scalar triple product of $\vec a_2-\vec a_1$, $\vec b_1$, $\vec b_2$ vanishes -- geometrically, this says the four points/directions all lie in one plane (the lines are coplanar). Written out in Cartesian form for lines $\dfrac{x-x_1}{a_1}=\dfrac{y-y_1}{b_1}=\dfrac{z-z_1}{c_1}$ and $\dfrac{x-x_2}{a_2}=\dfrac{y-y_2}{b_2}=\dfrac{z-z_2}{c_2}$, this same condition becomes the determinant equation

$\begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}=0.$ …