Answer

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**Hint:**Start by simplifying the double cross product. You can use the formula, \[\vec a \times (\vec b \times \vec c) = (\vec a.\vec c)\vec b - (\vec a.\vec b)\vec c\]. And then use that if the two components of the box product is equal then, its value will be zero.

**Complete step by step solution:**We have, given, \[(\vec d + \vec a)[\vec a \times (\vec b \times (\vec c \times \vec d))]\]

\[ = (\vec d + \vec a)[\vec a \times (\vec b.\vec d)\vec c - (\vec b.\vec c)\vec d]\]

As, \[\vec a \times (\vec b \times \vec c) = (\vec a.\vec c)\vec b - (\vec a.\vec b)\vec c\]

Now, again doing cross product, with the\[\vec a\]we get,

\[ = (\vec d + \vec a)[(\vec b.\vec d)(\vec a \times \vec c) - (\vec b.\vec c)(\vec a \times \vec d)]\]

Next, we do dot product with \[\vec a\]and \[\vec d\] one by one,

\[ = (\vec b.\vec d)[\vec d.(\vec a \times \vec c) + \vec a.(\vec a \times \vec c)] - (\vec b.\vec c)[\vec d.(\vec a \times \vec d) + \vec a.(\vec a \times \vec d)]\]

\[ = (\vec b.\vec d)[[\vec d\vec a\vec c] + [\vec a\vec a\vec c] - (\vec b.\vec c)[[\vec d\vec a\vec d] + [\vec a\vec a\vec d]]\]

As per the definition of the box product that can be written.

Box product is specified as, The scalar triple product (also called the mixed product, box product, or triple scalar product) is defined as the dot product of one of the vectors with the cross product of the other two.

Now, we also know if two elements of the box product are equal then its value turns out to be zero.

So, we have,

\[(\vec b.\vec d)[[\vec d\vec a\vec c] + [\vec a\vec a\vec c] - (\vec b.\vec c)[[\vec d\vec a\vec d] + [\vec a\vec a\vec d]]\]

\[ = (\vec b.\vec d)[\vec d\vec a\vec c]\]

As, \[[\vec a\vec a\vec d]\]\[ = \]\[[\vec d\vec a\vec d]\]\[ = \]\[[\vec a\vec a\vec c]\]\[ = \]0 as they have two similar terms.

**So, we have our answer as option a. \[ = (\vec b.\vec d)[\vec d\vec a\vec c]\]**

**Note:**If in the box product two terms are equal then the value will be zero can be proved in this way,

We have,

\[[\vec a\vec b\vec c] = \vec a.(\vec b \times \vec c)\]

Now, if any two of them are equal, say \[\vec b = \vec c\]

Then,

\[[\vec a\vec b\vec b] = \vec a.(\vec b \times \vec b)\]

But now, the value of \[(\vec b \times \vec b)\]= 0, so we will have the value of the box product turn out to be zero.

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