The vector \[\widehat{i}\times \left( \overrightarrow{a}\times \widehat{i} \right)+\widehat{j}\times \left( \overrightarrow{a}\times \widehat{j} \right)+\widehat{k}\times \left( \overrightarrow{a}\times \widehat{k} \right)\] is equal to
(a) \[\overrightarrow{0}\]
(b) \[\overrightarrow{a}\]
(c) \[2\overrightarrow{a}\]
(d) None of these

Hint: First of all, assume, \[\overrightarrow{a}=x\widehat{i}+y\widehat{j}+z\widehat{k}\]. Now use the formula for vector triple product that is \[\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right),\left( \overrightarrow{a}.\overrightarrow{c} \right)b-\left( \overrightarrow{a}.\overrightarrow{b} \right)c\] and also use \[\widehat{i}.\widehat{i}=\widehat{j}.\widehat{j}=\widehat{k}.\widehat{k}=1;\widehat{i}.\widehat{j}=\widehat{j}.\widehat{k}=\widehat{k}.\widehat{i}=0\] to find the value of the given expression. Also separate the given expression and then solve individually each pat to avoid any confusion.

Complete step-by-step solution -
In this question, we have to find the value of the vector
\[\widehat{i}\times \left( \overrightarrow{a}\times \widehat{i} \right)+\widehat{j}\times \left( \overrightarrow{a}\times \widehat{j} \right)+\widehat{k}\times \left( \overrightarrow{a}\times \widehat{k} \right)\]
First of all, let us consider the expression given in the question.
\[E=\widehat{i}\times \left( \overrightarrow{a}\times \widehat{i} \right)+\widehat{j}\times \left( \overrightarrow{a}\times \widehat{j} \right)+\widehat{k}\times \left( \overrightarrow{a}\times \widehat{k} \right)\]
Let us take the value of \[\overrightarrow{a}=x\widehat{i}+y\widehat{j}+z\widehat{k}\].
Let us assume,
\[\begin{align}
  & \overrightarrow{P}=\widehat{i}\times \left( \overrightarrow{a}\times \widehat{i} \right) \\
 & \overrightarrow{Q}=\widehat{j}\times \left( \overrightarrow{a}\times \widehat{j} \right) \\
 & \overrightarrow{R}=\widehat{k}\times \left( \overrightarrow{a}\times \widehat{k} \right) \\
\end{align}\]
So, we get the above expression as,
\[E=\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}....\left( i \right)\]
We know that according to the vector triple product,
\[\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{a}.\overrightarrow{b} \right)\overrightarrow{c}....\left( ii \right)\]
By using this, let us find the value of vectors P, Q, and R.
Let us find the value of vector P.
\[\overrightarrow{P}=\widehat{i}\times \left( \overrightarrow{a}\times \widehat{i} \right)\]
By substituting the value of \[\overrightarrow{a}=\widehat{i},\overrightarrow{b}=\overrightarrow{a},\overrightarrow{c}=\widehat{i}\]in equation (ii), we get,
\[\overrightarrow{P}=\left( \widehat{i}.\widehat{i} \right)\overrightarrow{a}-\left( \widehat{i}.\overrightarrow{a} \right)\widehat{i}\]
 By substituting, \[\overrightarrow{a}=x\widehat{i}+y\widehat{j}+z\widehat{k}\], we get,
\[\overrightarrow{P}=\left( \widehat{i}.\widehat{i} \right)\left[ x\widehat{i}+y\widehat{j}+z\widehat{k} \right]-\left[ \widehat{i}.\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right) \right]\widehat{i}\]
We know that,
\[\begin{align}
  & \left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right).\left( p\widehat{i}+q\widehat{j}+r\widehat{k} \right)=ap+bq+cr \\
 & \widehat{i}.\widehat{i}=\widehat{j}.\widehat{j}=\widehat{k}.\widehat{k}=1 \\
 & \widehat{i}.\widehat{j}=\widehat{j}.\widehat{k}=\widehat{k}.\widehat{i}=0 \\
\end{align}\]
By using these, we get,
\[\overrightarrow{P}=1\left[ x\widehat{i}+y\widehat{j}+z\widehat{k} \right]-\left[ x+0+0 \right]\widehat{i}\]
\[\overrightarrow{P}=x\widehat{i}+y\widehat{j}+z\widehat{k}-x\widehat{i}\]
\[\overrightarrow{P}=y\widehat{j}+z\widehat{k}....\left( iii \right)\]
Now, let us find the value of vector Q.
\[\overrightarrow{Q}=\widehat{j}\times \left( \overrightarrow{a}\times \widehat{j} \right)\]
By substituting the value of \[\overrightarrow{a}=\widehat{j},\overrightarrow{b}=\overrightarrow{a},\overrightarrow{c}=\widehat{j}\]in equation (ii), we get,
\[\overrightarrow{Q}=\left( \widehat{j}.\widehat{j} \right)\overrightarrow{a}-\left( \widehat{j}.\overrightarrow{a} \right)\widehat{j}\]
 By substituting, \[\overrightarrow{a}=x\widehat{i}+y\widehat{j}+z\widehat{k}\], we get,
\[\overrightarrow{Q}=\left( \widehat{j}.\widehat{j} \right)\left[ x\widehat{i}+y\widehat{j}+z\widehat{k} \right]-\left[ \widehat{j}.\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right) \right]\widehat{j}\]
\[\overrightarrow{Q}=1\left[ x\widehat{i}+y\widehat{j}+z\widehat{k} \right]-\left[ 0+y+0 \right]\widehat{j}\]
\[\overrightarrow{Q}=x\widehat{i}+y\widehat{j}+z\widehat{k}-y\widehat{j}\]
\[\overrightarrow{Q}=x\widehat{i}+z\widehat{k}....\left( iv \right)\]
Now, let us find the value of vector R.
\[\overrightarrow{R}=\widehat{k}\times \left( \overrightarrow{a}\times \widehat{k} \right)\]
By substituting the value of \[\overrightarrow{a}=\widehat{k},\overrightarrow{b}=\overrightarrow{a},\overrightarrow{c}=\widehat{k}\]in equation (ii), we get,
\[\overrightarrow{R}=\left( \widehat{k}.\widehat{k} \right)\overrightarrow{a}-\left( \widehat{k}.\overrightarrow{a} \right)\widehat{k}\]
 By substituting, \[\overrightarrow{a}=x\widehat{i}+y\widehat{j}+z\widehat{k}\], we get,
\[\overrightarrow{R}=\left( \widehat{k}.\widehat{k} \right)\left[ x\widehat{i}+y\widehat{j}+z\widehat{k} \right]-\left[ \widehat{k}.\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right) \right]\widehat{k}\]
\[\overrightarrow{R}=1\left[ x\widehat{i}+y\widehat{j}+z\widehat{k} \right]-\left[ 0+0+z \right]\widehat{k}\]
\[\overrightarrow{R}=x\widehat{i}+y\widehat{j}+z\widehat{k}-z\widehat{k}\]
\[\overrightarrow{R}=x\widehat{i}+y\widehat{j}....\left( v \right)\]
Now, by substituting the values of vectors P, Q and R from equation (iii), (iv) and (v), we get,
\[E=y\widehat{j}+z\widehat{k}+x\widehat{i}+z\widehat{k}+x\widehat{i}+y\widehat{j}\]
\[E=2x\widehat{i}+2y\widehat{j}+2z\widehat{k}\]
\[E=2\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right)\]
By replacing \[x\widehat{i}+y\widehat{j}+z\widehat{k}=\overrightarrow{a}\], we get,
\[E=2\overrightarrow{a}\]
Hence, we get the value of the given expression as \[2\overrightarrow{a}\].
Hence, the option (c) is the right answer.
\[\]

Note: In these types of questions, whenever the value of a certain vector is not given, we should always consider it as a general vector that is \[x\widehat{i}+y\widehat{j}+z\widehat{k}\]. Also, note that the vector triple product of 3 vectors is not associative that is, \[\left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \overrightarrow{c}\ne \overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)\] because
\[\begin{align}
  & \left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \overrightarrow{c}=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{b}.\overrightarrow{c} \right)\overrightarrow{a} \\
 & \overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{a}.\overrightarrow{b} \right)\overrightarrow{c} \\
\end{align}\]
So, it is advisable to apply the formula carefully to avoid any mistakes.