If the vectors \[2\widehat{i}-3\widehat{j}+4\widehat{k}\], \[\widehat{i}+2\widehat{j}-\widehat{k}\] and \[x\widehat{i}-\widehat{j}+2\widehat{k}\] are coplanar, then \[x=\]
A. \[\dfrac{8}{5}\]
B. \[\dfrac{5}{8}\]
C. \[0\]
D. \[1\]

Hint: In the above question, we are given three vectors that are coplanar and we have to calculate the coefficient of the direction cosines of one of the vectors. We can calculate the value of the coefficient very easily if we know the concept of determinant calculation using the matrix method in the case of a coplanar matrix.

Formula used: Scalar triple product of three vectors:
We have the vectors \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] as
\[\begin{align}
  & \overrightarrow{a}={{a}_{1}}\overrightarrow{i}+{{a}_{2}}\overrightarrow{j}+{{a}_{3}}\overrightarrow{k} \\
 & \overrightarrow{b}={{b}_{1}}\overrightarrow{i}+{{b}_{2}}\overrightarrow{j}+{{b}_{3}}\overrightarrow{k} \\
 & \overrightarrow{c}={{c}_{1}}\overrightarrow{i}+{{c}_{2}}\overrightarrow{j}+{{c}_{3}}\overrightarrow{k} \\
\end{align}\]
Then, the triple product is calculated by,
\[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|\]
Thus, the volume of a 3D structure is calculated by $V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]$

Complete step by step solution: Here, we are given three vectors that are coplanar which means they lie in the same plane.
To get the solution to the given question, we should remember that the determinant of the coefficient matrix of the direction cosines of the given matrix is equal to zero.
In order to calculate the value of the unknown coefficient $x$, we have to write the coefficients of the direction cosines of these vectors in a matrix.
The given vectors are:
\[\begin{align}
  & \overrightarrow{a}=\widehat{i}+2\widehat{j}-\widehat{k} \\
 & \overrightarrow{b}=\widehat{i}+2\widehat{j}-\widehat{k} \\
 & \overrightarrow{c}=x\widehat{i}-\widehat{j}+2\widehat{k} \\
\end{align}\]
Then, the scalar triple product of these vectors is calculated by,
\[\begin{align}
  & [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right| \\
 & \text{ }=\left| \begin{matrix}
   2 & -3 & 4 \\
   1 & 2 & -1 \\
   x & -1 & 2 \\
\end{matrix} \right| \\
 & \text{ }=2(4-1)+3(2+x)+4(-1-2x) \\
 & \text{ }=6+6+3x-4-8x \\
 & \text{ }=8-5x \\
\end{align}\]
We know from the property of the coplanarity matrix that its determinant is equal is zero.
So,
\[\begin{align}
  & [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=0 \\
 & \Rightarrow 8-5x=0 \\
 & \Rightarrow 5x=8 \\
 & \therefore x=\dfrac{8}{5} \\
\end{align}\]

Thus, Option (A) is correct.

Note:Here we may go wrong with the vector identities and scalar triple product. Here are the simple formulas used for solving the given vector. By applying appropriate vector products, the given vector equation is evaluated.