A force, $\overrightarrow{F}=3\widehat{i}+c\widehat{j}+2\widehat{k}$N acting on a particle causes a displacement of, $\overrightarrow{S}=-4\widehat{i}+2\widehat{j}-3\widehat{k}$m. If the work done is 6 Joule, the value of ‘c’ is:
(A) 0
(B) 1
(C) 12
(D) 6

Hint: In this problem, we have been given the force vector and the displacement vector. Now, the work done by a force over a certain displacement is given by the scalar dot product of force vector and displacement vector. We shall use this concept to find the Y-component of the force vector as the net work done by this force has been given to us.

Complete answer:
It has been given to us in the problem that:
The force vector is given by:
$\Rightarrow \overrightarrow{F}=(3\widehat{i}+c\widehat{j}+2\widehat{k})N$
And, the displacement vector is given by:
 $\Rightarrow \overrightarrow{S}=(-4\widehat{i}+2\widehat{j}-3\widehat{k})m$
Since, the work done by a force over a certain displacement is given by the scalar product of force vector and displacement vector. Therefore, mathematically the above statement could be written as follows:
$\Rightarrow W=\overrightarrow{F}.\overrightarrow{S}$ [Let this expression be equation number (1)]
Where, W is the total work done. Since, the dot product of two vectors is a scalar quantity, hence work done is also a scalar quantity.
Putting the values of force vector, displacement vector and total work done in equation number (1), we get:
$\begin{align}
  & \Rightarrow W=\left( 3\widehat{i}+c\widehat{j}+2\widehat{k} \right).\left( -4\widehat{i}+2\widehat{j}-3\widehat{k} \right) \\
 & \Rightarrow 6=-12+2c-6 \\
 & \Rightarrow 2c=24 \\
 & \therefore c=12 \\
\end{align}$
Therefore, the net force vector could be written as:
$\Rightarrow \overrightarrow{F}=\left( 3\widehat{i}+12\widehat{j}+2\widehat{k} \right)$
Hence, the value of constant ‘c’ comes out to be 12.

Hence, option (C) is the correct option.

Note:
We should be aware of the basic mathematical definition of different terms like work, power, energy, etc. in physics. Also, while performing the dot product, we used the property that the dot product of any two perpendicular vectors is zero and the dot product of two parallel vectors is equal to their product.