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Hint: For solving this problem, first we solve the inner cross product of the three involved for non-coplanar vectors. Now, expanding the dot product of the given expression, we obtain a simplified result in the scalar triple product of vectors.

Vectors are represented by directed line segments such that the length of the line segment is the magnitude of the vector and direction of the arrow marked is the direction of the vector. Three vectors are said to be non-coplanar, if their support lines are not parallel to the same plane or they cannot be expressed as $\overrightarrow{R}=x\overrightarrow{A}+y\overrightarrow{B}+z\overrightarrow{C}$.

According to the problem statement, we are given $\left( A+B+C \right)\cdot \left[ \left( A+B \right)\times \left( A+C \right) \right]$

Now, first expanding the cross product of the expression, we get

$\left( A+B+C \right)\cdot \left[ A\times A+A\times C+B\times A+B\times C \right]$

By using the identity $A\times A=0$, we simplify the expression as

$\left( A+B+C \right)\cdot \left[ A\times C+B\times A+B\times C \right]$

Now, on expanding the dot product and by using the identity that scalar triple product involving two similar vectors is zero, we get

\[\begin{align}

& A\cdot \left( A\times C \right)+A\cdot \left( B\times A \right)+A\cdot \left( B\times C \right)+B\cdot \left( A\times C \right)+B\cdot \left( B\times A \right)+B\cdot \left( B\times C \right)+C\cdot \left( A\times C \right) \\

& +C\cdot \left( B\times A \right)+C\cdot \left( B\times C \right) \\

& \Rightarrow C\cdot \left( B\times A \right)+B\cdot \left( A\times C \right)+A\cdot \left( B\times C \right) \\

\end{align}\]

Now, by using the property $C\cdot \left( B\times A \right)=-\left[ A,B,C \right],B\cdot \left( A\times C \right)=-\left[ A,B,C \right]\text{ and}\,A.\left( B\times C \right)=\left[ A,B,C \right]$, we get

\[\begin{align}

& \Rightarrow -\left[ A,B,C \right]-\left[ A,B,C \right]+\left[ A,B,C \right] \\

& \Rightarrow -\left[ A,B,C \right] \\

\end{align}\]

Therefore option (d) is correct.

Note: The key concept involved in solving this problem is the knowledge of scalar triple product of non-collinear vectors. Students must remember that the vectors are non collinear so their scalar triple product cannot be zero. Also, the negative sign due to clockwise rotation of vectors must be taken into account.

__Complete step-by-step solution -__Vectors are represented by directed line segments such that the length of the line segment is the magnitude of the vector and direction of the arrow marked is the direction of the vector. Three vectors are said to be non-coplanar, if their support lines are not parallel to the same plane or they cannot be expressed as $\overrightarrow{R}=x\overrightarrow{A}+y\overrightarrow{B}+z\overrightarrow{C}$.

According to the problem statement, we are given $\left( A+B+C \right)\cdot \left[ \left( A+B \right)\times \left( A+C \right) \right]$

Now, first expanding the cross product of the expression, we get

$\left( A+B+C \right)\cdot \left[ A\times A+A\times C+B\times A+B\times C \right]$

By using the identity $A\times A=0$, we simplify the expression as

$\left( A+B+C \right)\cdot \left[ A\times C+B\times A+B\times C \right]$

Now, on expanding the dot product and by using the identity that scalar triple product involving two similar vectors is zero, we get

\[\begin{align}

& A\cdot \left( A\times C \right)+A\cdot \left( B\times A \right)+A\cdot \left( B\times C \right)+B\cdot \left( A\times C \right)+B\cdot \left( B\times A \right)+B\cdot \left( B\times C \right)+C\cdot \left( A\times C \right) \\

& +C\cdot \left( B\times A \right)+C\cdot \left( B\times C \right) \\

& \Rightarrow C\cdot \left( B\times A \right)+B\cdot \left( A\times C \right)+A\cdot \left( B\times C \right) \\

\end{align}\]

Now, by using the property $C\cdot \left( B\times A \right)=-\left[ A,B,C \right],B\cdot \left( A\times C \right)=-\left[ A,B,C \right]\text{ and}\,A.\left( B\times C \right)=\left[ A,B,C \right]$, we get

\[\begin{align}

& \Rightarrow -\left[ A,B,C \right]-\left[ A,B,C \right]+\left[ A,B,C \right] \\

& \Rightarrow -\left[ A,B,C \right] \\

\end{align}\]

Therefore option (d) is correct.

Note: The key concept involved in solving this problem is the knowledge of scalar triple product of non-collinear vectors. Students must remember that the vectors are non collinear so their scalar triple product cannot be zero. Also, the negative sign due to clockwise rotation of vectors must be taken into account.

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