Dot product of two vectors `overset(rarr)A` and `overset(rarr)B` is defined as `overset(rarr)A.overset(rarr)B=AB cos phi` , where `phi` is angle between them when they are drawn with tails coinciding. For any two vectors . This means `ovsert(rarr)A . overset(rarr)B=overset(rarr)B. overset(rarr)A` that . The scalar product obeys the commutative law of multiplication, the order of the two vectors does not matter. The vector product of two vectors `overset(rarr)A` and `overset(rarr)B` also called the cross product, is denoted by `overset(rarr)A xx overset(rarr)B` . As the name suggests, the vector product is itself a vector. `overset(rarr)C=overset(rarr)A xx overset(rarr)B` then `C=AB sin theta` ,
For non zero vectors `overset(rarr)A, overset(rarr)B, overset(rarr)C,|(overset(rarr)Axxoverset(rarr)B).overset(rarr)C|=|overset(rarr)A||overset(rarr)B||overset(rarr)C|` holds if and only if

Options

A. `overset(rarr)A.overset(rarr)B=0,overset(rarr)B.overset(rarr)C=0`
B. `overset(rarr)B.overset(rarr)C=0,overset(rarr)C.overset(rarr)A=0`
C. `overset(rarr)C.overset(rarr)A=0,overset(rarr)A.overset(rarr)B=0`
D. `overset(rarr)A.overset(rarr)B=overset(rarr)B.overset(rarr)C=overset(rarr)C.overset(rarr)A=0`

Explanation

To solve the problem, we need to analyze the given expression involving the cross product and the dot product of vectors. Let's break it down step by step.

### Step 1: Understand the Given Expression
We have the expression:
\[ |(\overset{\rarr}{A} \times \overset{\rarr}{B}) \cdot \overset{\rarr}{C}| = |\overset{\rarr}{A}| |\overset{\rarr}{B}| |\overset{\rarr}{C}| \]
This expression states that the magnitude of the dot product of the vector resulting from the cross product of vectors A and B with vector C equals the product of the magnitudes of vectors A, B, and C.

### Step 2: Expand the Left Side
The left side can be expanded using the definition of the cross product and the dot product:
\[ \overset{\rarr}{C} = \overset{\rarr}{A} \times \overset{\rarr}{B} \]
The magnitude of the cross product is given by:
\[ |\overset{\rarr}{A} \times \overset{\rarr}{B}| = |\overset{\rarr}{A}| |\overset{\rarr}{B}| \sin \phi \]
where \(\phi\) is the angle between vectors A and B.

Now, we can write:
\[ |(\overset{\rarr}{A} \times \overset{\rarr}{B}) \cdot \overset{\rarr}{C}| = |\overset{\rarr}{A} \times \overset{\rarr}{B}| |\overset{\rarr}{C}| \cos \theta \]
where \(\theta\) is the angle between the vector \((\overset{\rarr}{A} \times \overset{\rarr}{B})\) and vector C.

### Step 3: Set Up the Equation
Substituting the expression for the magnitude of the cross product into the equation gives:
\[ |\overset{\rarr}{A}| |\overset{\rarr}{B}| \sin \phi |\overset{\rarr}{C}| \cos \theta = |\overset{\rarr}{A}| |\overset{\rarr}{B}| |\overset{\rarr}{C}| \]

### Step 4: Simplify the Equation
We can cancel out the common terms (assuming none of the vectors are zero):
\[ \sin \phi \cos \theta = 1 \]

### Step 5: Analyze the Condition
The equation \(\sin \phi \cos \theta = 1\) implies:
1. \(\sin \phi = 1\) and \(\cos \theta = 1\)

From \(\sin \phi = 1\), we find that:
\[ \phi = 90^\circ \]
This means that vectors A and B are perpendicular to each other.

From \(\cos \theta = 1\), we find that:
\[ \theta = 0^\circ \]
This means that vector C is parallel to the vector resulting from the cross product of A and B.

### Step 6: Conclusion
Thus, for the expression to hold true, we conclude that:
- Vectors A and B are perpendicular (\(\phi = 90^\circ\)).
- Vector C is parallel to the vector resulting from the cross product of A and B (\(\theta = 0^\circ\)).

### Final Answer
The condition holds true if and only if:
- \(\overset{\rarr}{A} \perp \overset{\rarr}{B}\) and \(\overset{\rarr}{C} \parallel (\overset{\rarr}{A} \times \overset{\rarr}{B})\).