$\begin{vmatrix} 1 & 1 & -1 \\ p & 2 & 1 \\ 4 & 2p & 7 \end{vmatrix} = 0$
Step-by-Step Solution
Key Concept: When three lines are concurrent, the determinant formed by their coefficients equals zero.
Expanding the determinant: $1(14 - 2p) - 1(7p - 4) - 1(2p^2 - 8) = 0$. This simplifies to $14 - 2p - 7p + 4 - 2p^2 + 8 = 0$, giving $2p^2 + 9p - 26 = 0$. Using the quadratic formula: $p = \frac{-9 \pm \sqrt{81 + 208}}{4} = \frac{-9 \pm 17}{4}$, so $p = 2$ or $p = -\frac{13}{2}$. Since $p = 2$ is rejected because $p = 2$ all 3 lines becomes parallel, hence $p = -\frac{13}{2}$.
Correct Answer: -13/2