The given determinant $\left| {\begin{array}{*{20}{c}}
  {103}&{115}&{114} \\
  {111}&{108}&{106} \\
  {104}&{113}&{116}
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
  {113}&{116}&{104} \\
  {108}&{106}&{111} \\
  {115}&{114}&{103}
\end{array}} \right|$ is equal to
A.A positive number
B.A negative number
C.Zero
D.None of these

Hint: We can take the 2nd matrix and interchange the 1st and 2nd column by changing the sign of the determinant. Then we can interchange the 1st and 3rd column by changing the sign of the determinant. After that we can interchange the 1st and 3rd rows by changing the sign of the determinant. Then we will get the same matrix as that of the 1st matrix. After simplification, we will obtain the required answer.

Complete step-by-step answer:
We have 2 determinants,
 $\left| {\begin{array}{*{20}{c}}
  {103}&{115}&{114} \\
  {111}&{108}&{106} \\
  {104}&{113}&{116}
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
  {113}&{116}&{104} \\
  {108}&{106}&{111} \\
  {115}&{114}&{103}
\end{array}} \right|$
Let $A = \left| {\begin{array}{*{20}{c}}
  {103}&{115}&{114} \\
  {111}&{108}&{106} \\
  {104}&{113}&{116}
\end{array}} \right|$ and $B = \left| {\begin{array}{*{20}{c}}
  {113}&{116}&{104} \\
  {108}&{106}&{111} \\
  {115}&{114}&{103}
\end{array}} \right|$
On observing the matrix, we can see that the elements in A are the same as the elements in B.
We can consider B, we can see that the elements in the 1st column of B are the elements of 2nd column of A. So, we can do the column operation ${C_1} \leftrightarrow {C_2}$ . We know that interchanging 2 rows or columns changes the sign of the determinant. Therefore, B becomes
 $B = \left| {\begin{array}{*{20}{c}}
  {113}&{116}&{104} \\
  {108}&{106}&{111} \\
  {115}&{114}&{103}
\end{array}} \right|$
 ${C_1} \leftrightarrow {C_2}$
 $ \Rightarrow B = - \left| {\begin{array}{*{20}{c}}
  {116}&{113}&{104} \\
  {106}&{108}&{111} \\
  {114}&{115}&{103}
\end{array}} \right|$
Now we can interchange the 1st and 3rd column by changing the sign.
 ${C_1} \leftrightarrow {C_3}$
 $ \Rightarrow B = \left| {\begin{array}{*{20}{c}}
  {104}&{113}&{116} \\
  {111}&{108}&{106} \\
  {103}&{115}&{114}
\end{array}} \right|$
Now on comparing with A, we can see that by interchanging the 1st and 3rd row of B, we can obtain the matrix A.
 ${R_1} \leftrightarrow {R_3}$
  $ \Rightarrow B = - \left| {\begin{array}{*{20}{c}}
  {103}&{115}&{114} \\
  {111}&{108}&{106} \\
  {104}&{113}&{116}
\end{array}} \right|$
 $ \Rightarrow B = - A$
Thus, the determinants will become,
 \[A + B\]
 $ = \left| {\begin{array}{*{20}{c}}
  {103}&{115}&{114} \\
  {111}&{108}&{106} \\
  {104}&{113}&{116}
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
  {113}&{116}&{104} \\
  {108}&{106}&{111} \\
  {115}&{114}&{103}
\end{array}} \right|$
As we have \[B = - A\] , so we get,
 $ = \left| {\begin{array}{*{20}{c}}
  {103}&{115}&{114} \\
  {111}&{108}&{106} \\
  {104}&{113}&{116}
\end{array}} \right| - \left| {\begin{array}{*{20}{c}}
  {103}&{115}&{114} \\
  {111}&{108}&{106} \\
  {104}&{113}&{116}
\end{array}} \right|$
As the 2 determinants are same so they cancel out,
 $ = 0$
Therefore, the value of the given expression is zero.
So, the correct answer is option C.

Note: Determinant of a matrix is a number related to a square matrix. The operations we did on the matrix are called elementary row and column operations. It is the property of determinants that the determinant of a matrix will be multiplied with -1 if we do one column or row exchange. The operations we did to solve this problem need not to be unique. We can use any order of exchange to get the determinant of the 2nd matrix as negative of the determinant of the 1st matrix.