Solve the given determinant for x:
$\left| \begin{align}
  x-2 \;\;\;\; \;\;\;\; & 2x-3 & 3x-4 \\
 x-4 \;\;\;\; \;\;\;\; & 2x-9 & 3x-16 \\
 x-8 \;\;\;\; \;\;\;\; & 2x-27 & 3x-64 \\
\end{align} \right|=0$
(A). $\dfrac{7}{4}$
(B). $4$
(C). $\dfrac{28}{13}$
(D). $\dfrac{14}{9}$

Hint: Apply the transformations on row 1 and row 2 of the given determinant. The transformation that we are going to do is to transform row 1 by subtracting row 2 from row 1 and to transform row 2 by subtracting row 3 from row 2 then expand along row 1 and then solve the equation to find the value of x.

Complete step-by-step solution -
The determinant equation which is given above:
$\left| \begin{align}
 x-2 \;\;\;\; \;\;\;\; & 2x-3 & 3x-4 \\
 x-4 \;\;\;\; \;\;\;\; & 2x-9 & 3x-16 \\
 x-8 \;\;\;\; \;\;\;\; & 2x-27 & 3x-64 \\
\end{align} \right|=0$
We are going to transform row 1 of the above determinant by subtracting row 2 from row 1 as follows:
$\left| \begin{align}
  2 \;\;\;\; \;\;\;\;& 6 & 12\\
 x-4 \;\;\;\; \;\;\;\;& 2x-9 & 3x-16 \\
 x-8 \;\;\;\; \;\;\;\;& 2x-27 & 3x-64 \\
\end{align} \right|=0$
Now, we are going to transform row 2 by subtracting row 3 from row 2 as follows:
$\left| \begin{align}
   2 \;\;\;\; \;\;\;\;& 6 & 12 \\
   4 \;\;\;\; \;\;\;\;& 18 & 48 \\
  x-8 \;\;\;\; \;\;\;\;& 2x-27 & 3x-64 \\
\end{align} \right|=0$
Expanding the above determinant along the first row we get,
$\begin{align}
  & 2\left( 18\left( 3x-64 \right)-48\left( 2x-27 \right) \right)-6\left( 4\left( 3x-64 \right)-48\left( x-8 \right) \right)+12\left( 4\left( 2x-27 \right)-18\left( x-8 \right) \right)=0 \\
 & \Rightarrow 2\left( 54x-1152-96x+1296 \right)-6\left( 12x-256-48x+384 \right)+12\left( 8x-108-18x+144 \right)=0 \\
 & \Rightarrow 2\left( -42x+144 \right)-6\left( -36x+128 \right)+12\left( -10x+36 \right)=0 \\
 & \Rightarrow -84x+288+216x-768-120x+432=0 \\
 & \Rightarrow 12x-48=0 \\
 & \Rightarrow x=4 \\
\end{align}$
From the above solution, we get the value of x is equal to 4.
Hence, the correct option is (b).

Note: You can check whether the value of x that we have got is correct or not by substituting the value of x in the given determinant and see whether by plugging the value of x will make the determinant value 0 or not.
Substituting the value of x = 4 in the given determinant we get,
$\left| \begin{align}
  x-2 \;\;\;\; \;\;\;\;& 2x-3 & 3x-4 \\
  x-4 \;\;\;\; \;\;\;\;& 2x-9 & 3x-16 \\
 x-8 \;\;\;\; \;\;\;\;& 2x-27 & 3x-64 \\
\end{align} \right| $
 $ \Rightarrow \left| \begin{align}
   2 \;\;\;\; \;\;\;\;& 5 & 8 \\
   0 \;\;\;\; \;\;\;\;& -1 & -4 \\
  -4 \;\;\;\; \;\;\;\;& -19 & -52 \\
\end{align} \right| \\
$
Now, expanding the above determinant along second row to get the value of the determinant we get,
$\begin{align}
  & 0-1\left( 2\left( -52 \right)+32 \right)+4\left( -38+20 \right) \\
 & =-1\left( -104+32 \right)+4\left( -18 \right) \\
 & =72-72 = 0 \\
\end{align}$