$\left| \begin{matrix}

49 & 1 & 6  \\

39 & 7 & 4  \\

26 & 2 & 3  \\

\end{matrix} \right|$

Let $\vartriangle =\left| \begin{matrix}

49 & 1 & 6  \\

39 & 7 & 4  \\

26 & 2 & 3  \\

\end{matrix} \right|$

Now by applying column operation, ${{C}_{1}}\to {{C}_{1}}-8{{C}_{3}}$, we get

$\vartriangle =\left| \begin{matrix}

1 & 1 & 6  \\

7 & 7 & 4  \\

2 & 2 & 3  \\

\end{matrix} \right|$

As, ${{C}_{1}}={{C}_{2}}$, hence determinant Is zero.