Question

$| \begin{array}{cccccccccc} 1 & 1 & 2 \\ 2 & 2 & 4 \\ 3 & 5 & 6 \end{array} | =$

1st Way :-

Expanding along 1st row, we get

\begin{array}{l} \begin{aligned} | \begin{array}{cccccccccc} 1 & 1 & 2 \\ 2 & 2 & 4 \\ 3 & 5 & 6 \end{array} | & = & 1 \cdot | \begin{array}{cccccccccc} 2 & 4 \\ 5 & 6 \end{array} | - 1 \cdot | \begin{array}{cccccccccc} 2 & 4 \\ 3 & 6 \end{array} | + 2 \cdot | \begin{array}{cccccccccc} 2 & 2 \\ 3 & 5 \end{array} | \\ = & 1 \cdot (2 \cdot 6 - 4 \cdot 5) - 1 \cdot (2 \cdot 6 - 4 \cdot 3) \\ + 2 \cdot (2 \cdot 5 - 2 \cdot 3) \\ = & (12 - 20) - (12 - 12) + 2 \cdot (10 - 6) \\ = & - 8 - 0 + 2 \cdot 4 \\ = & - 8 + 8 \\ = & 0 \cdot \end{aligned} \end{array}

2nd Way :-

We know that in determinants if two rows or two columns are the same then determinants is zero.

Here in given determinant if we take 2 common from 2nd row then, we get

\begin{array}{l} \begin{aligned} | \begin{array}{cccccccccc} 1 & 1 & 2 \\ 2 & 2 & 4 \\ 3 & 5 & 6 \end{array} | & = & 2 | \begin{array}{cccccccccc} 1 & 1 & 2 \\ 1 & 1 & 2 \\ 3 & 5 & 6 \end{array} | \end{aligned} \end{array}

Now, we see that 1st and 2nd rows are the same hence determinant is zero

\begin{aligned} | \begin{array}{cccccccccc} 1 & 1 & 2 \\ 2 & 2 & 4 \\ 3 & 5 & 6 \end{array} | & = & 0 \cdot \end{aligned}

3rd Way :-

We know that in determinants if two rows or two columns are the same then determinants is zero.

Here in given determinant if we take 2 common from 3rd column then, we get

\begin{array}{l} \begin{aligned} | \begin{array}{cccccccccc} 1 & 1 & 2 \\ 2 & 2 & 4 \\ 3 & 5 & 6 \end{array} | & = & 2 | \begin{array}{cccccccccc} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 5 & 3 \end{array} | \end{aligned} \end{array}

Now, we see that 1st and 2nd rows are the same also 1st column and 3rd colunm are the same hence determinant is zero

\begin{array}{l} \begin{array}{ccc} | \begin{array}{cccccccccc} 1 & 1 & 2 \\ 2 & 2 & 4 \\ 3 & 5 & 6 \end{array} | & = & 0 \cdot \end{array} \end{array}

So, correct answer is option-3, $\to 0$