Q.Find the value of $k$ for which the lines $x + y - 5 = 0$, $\ x - y - 1 = 0$ and $\ 2x + ky - 12 = 0$ are concurrent.

Three lines are concurrent exactly when the $3\times3$ determinant of their coefficients and constants (each line written as $ax+by+c=0$) is zero.

Each line is already in the form $ax+by+c=0$. The three lines are concurrent if and only if $\begin{vmatrix} 1 & 1 & -5 \\ 1 & -1 & -1 \\ 2 & k & -12 \end{vmatrix} = 0.$

Expand along the first row: $1\begin{vmatrix} -1 & -1 \\ k & -12 \end{vmatrix} - 1\begin{vmatrix} 1 & -1 \\ 2 & -12 \end{vmatrix} + (-5)\begin{vmatrix} 1 & -1 \\ 2 & k \end{vmatrix}.$

Each minor: $\begin{vmatrix} -1 & -1 \\ k & -12 \end{vmatrix} = (-1)(-12)-(-1)(k) = 12+k,\quad \begin{vmatrix} 1 & -1 \\ 2 & -12 \end{vmatrix} = -12-(-2) = -10,\quad \begin{vmatrix} 1 & -1 \\ 2 & k \end{vmatrix} = k-(-2) = k+2.$ …

The commonest mistake is reading the coefficients into the determinant without first moving the constant to the left side as ax + by + c = 0 - a line written as x + y = 5 must be entered with c = …