How can I quickly find the determinant of this matrix $$
        \begin{vmatrix}
        14 & 2 & 1 & 3\\
        31 & 4 & 5 & 6\\
        26 & 3 & 7 & 4\\
        10 & 1 & 3 & 2\\
        \end{vmatrix}
       =
        \begin{vmatrix}
        5\cdot2+1+3 & 2 & 1 & 3\\
        5\cdot4+5+6 & 4 & 5 & 6\\
        5\cdot3+7+4 & 3 & 7 & 4\\
        5\cdot1+3+2 & 1 & 3 & 2\\
        \end{vmatrix}$$$$
       =
        \begin{vmatrix}
        5\cdot2 & 2 & 1 & 3\\
        5\cdot4 & 4 & 5 & 6\\
        5\cdot3 & 3 & 7 & 4\\
        5\cdot1 & 1 & 3 & 2\\
        \end{vmatrix}
        +
        \begin{vmatrix}
        1 & 2 & 1 & 3\\
        5 & 4 & 5 & 6\\
        7 & 3 & 7 & 4\\
        3 & 1 & 3 & 2\\
        \end{vmatrix}
        +
        \begin{vmatrix}
        3 & 2 & 1 & 3\\
        6 & 4 & 5 & 6\\
        4 & 3 & 7 & 4\\
        2 & 1 & 3 & 2\\
        \end{vmatrix}
$$
However I am not able to proceed beyond. The answer given is zero. Is there any simple determinant property that I am not able to guess?

The two last determinants in your right side are zero as they both have two equal columns, so you're left with:
$$\begin{vmatrix}
        14 & 2 & 1 & 3\\
        31 & 4 & 5 & 6\\
        26 & 3 & 7 & 4\\
        10 & 1 & 3 & 2\\
        \end{vmatrix}=  \begin{vmatrix}
        5\cdot2 & 2 & 1 & 3\\
        5\cdot4 & 4 & 5 & 6\\
        5\cdot3 & 3 & 7 & 4\\
        5\cdot1 & 1 & 3 & 2\\
        \end{vmatrix}=  5\begin{vmatrix}
        2 & 2 & 1 & 3\\
        4 & 4 & 5 & 6\\
        3 & 3 & 7 & 4\\
        1 & 1 & 3 & 2\\
        \end{vmatrix}=0
$$
Of course, much easier, short and clear, after you wrote your first equality sign, is to remark that the determinant is zero directly as
$$C_1=5C_2+C_3+C_4\;,\;\;C_i=i-\text{th column}$$
so that the matrix's rank isn't full and etc.
