If $\left| \begin{matrix}
   a-b-c & 2a & 2a \\
   2b & b-c-a & 2b \\
   2c & 2c & c-a-b \\
\end{matrix} \right|=\left( a+b+c \right){{\left( x+a+b+c \right)}^{2}},x\ne 0$ and $a+b+c\ne 0$ then $x$ is equal to
 (a) $-\left( a+b+c \right)$
(b) $2\left( a+b+c \right)$
(c) $abc$
(d) $-2\left( a+b+c \right)$

Hint: To find the value of $x$ we will find the determinant , we will first apply some row and column operation like ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}},\text{ }{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\text{ }and\text{ }{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}$ to simplify then we will expand the determinant to find the determinant. At last we will compare our Determinant with the one which was given to us originally to find the value of $x$ .

Complete step-by-step answer:
We are given that the Determinant of one matrix is given as $\left( a+b+c \right){{\left( x+a+b+c \right)}^{2}}$ we have been asked to find the value of $x$ .
So, we will first find the determinant of the matrix.
As we know row operators has no effect on Determinant, so we use row operation on the
Determinant to Simplify a bit.
Now, we have the determinant which is given as
 $\left| \begin{matrix}
   a-b-c & 2a & 2a \\
   2b & b-c-a & 2b \\
   2c & 2c & c-a-b \\
\end{matrix} \right|$
Now we will add all the rows.
That is ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}$ , so we will get,
 $\left| \begin{matrix}
   a+b+c & a+b+c & a+b+c \\
   2b & b-c-a & 2b \\
   2c & 2c & c-a-b \\
\end{matrix} \right|$
Now $a+b+c$ is common in the first row, so we will take it out from Row $1$ , so we get,
 $\left( a+b+c \right)\left| \begin{matrix}
   1 & 1 & 1 \\
   2b & b-c-a & 2b \\
   2c & 2c & c-a-b \\
\end{matrix} \right|$
Now we will subtract column 2 from column 1 and also we will subtract column 3 from column 1, that is, we will apply,
${{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\text{ }and\text{ }{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}$
So we will get as follows.
 \[\left( a+b+c \right)\left| \begin{matrix}
   1 & 0 & 0 \\
   2b & -\left( a+b+c \right) & 0 \\
   2c & 0 & -c-a-b \\
\end{matrix} \right|\]
Now we will expand along Row $1$ , so we will get
 \[=\left( a+b+c \right)\left[ 1\left( -\left( a+b+c \right)\left( -c-a-b \right) \right) \right]\]
Now, on simplifying the above expression, we will get,
 \[=\left( a+b+c \right){{\left( a+b+c \right)}^{2}}\]
So we get,
 $\left| \begin{matrix}
   a-b-c & 2a & 2a \\
   2b & b-c-a & 2b \\
   2c & 2c & c-a-b \\
\end{matrix} \right|=\left( a+b+c \right){{\left( a+b+c \right)}^{2}}$
On comparing this determinant with the one given to us we will get,
 $\left( a+b+c \right){{\left( x+a+b+c \right)}^{2}}=\left( a+b+c \right){{\left( a+b+c \right)}^{2}}$
On cancelling the like terms on both the sides, we will get,
 ${{\left( x+a+b+c \right)}^{2}}={{\left( a+b+c \right)}^{2}}$
 $\begin{align}
  & \Rightarrow \left( x+a+b+c \right)=a+b+c \\
 & \text{ }or \\
 & x+a+b+c=-\left( a+b+c \right) \\
\end{align}$
If we consider,
 $x+a+b+c=a+b+c$
Then we can write,
 $x=a+b+c-\left( a+b+c \right)$
 $x=0$
But $x\ne 0$ so this is not correct.
So we will consider,
 \[\begin{align}
  & x+a+b+c=-\left( a+b+c \right) \\
 & \Rightarrow x=-\left( a+b+c \right)-\left( a+b+c \right) \\
 & \text{ }x=-2\left( a+b+c \right) \\
\end{align}\]
So we will get a value of x, which is not equal to 0.
Therefore, the correct option is \[-2\left( a+b+c \right)\] option(d).

So, the correct answer is “Option d”.

Note: Remember while comparing the two Determinant we cannot simplify like,
 $\left( a+b+c \right){{\left( x+a+b+c \right)}^{2}}=\left( a+b+c \right){{\left( a+b+c \right)}^{2}}$
 $\Rightarrow {{\left( x+a+b+c \right)}^{2}}={{\left( a+b+c \right)}^{2}}$
Cancelling the squares on both the sides, we get,
 $\Rightarrow x+a+b+c=a+b+c$
After solving we will get,
 $x=0$ which is an incorrect option.
So while comparing squares we have to take both values of it.
That is,
 ${{\left( x+a+b+c \right)}^{2}}={{\left( a+b+c \right)}^{2}}$
And this will give us,
 $\begin{align}
  & \Rightarrow \left( x+a+b+c \right)=a+b+c \\
 & \text{ }or \\
 & x+a+b+c=-\left( a+b+c \right) \\
\end{align}$