Let a, b and c are three vectors hacing magnitude 1, 2 and 3 respectively satisfying the relation [a b c]=6. If `hat(d)` is a unit vector coplanar with b and c such that `b*hat(d)=1`, then evaluate `|(axxc)*d|^(2)+|(axxc)xxhat(d)|^(2)`.

Answer

To solve the problem step by step, we will evaluate the expression \( |(a \times c) \cdot d|^2 + |(a \times c) \times \hat{d}|^2 \).

### Step 1: Understand the given information
We have three vectors \( a, b, c \) with magnitudes:
- \( |a| = 1 \)
- \( |b| = 2 \)
- \( |c| = 3 \)

The scalar triple product \( [a \, b \, c] = a \cdot (b \times c) = 6 \).

### Step 2: Analyze the scalar triple product
The scalar triple product can also be expressed as:
\[ [a \, b \, c] = |a| \cdot |b| \cdot |c| \cdot \sin(\theta) \]
where \( \theta \) is the angle between \( a \) and the vector \( b \times c \). Since we know that \( |a| = 1 \), \( |b| = 2 \), and \( |c| = 3 \), we can write:
\[ 6 = 1 \cdot 2 \cdot 3 \cdot \sin(\theta) \implies \sin(\theta) = 1 \]
This means \( a \) is perpendicular to the plane formed by \( b \) and \( c \).

### Step 3: Find \( |a \times c| \)
Using the properties of cross products:
\[ |a \times c| = |a| \cdot |c| \cdot \sin(\phi) \]
where \( \phi \) is the angle between \( a \) and \( c \). Since \( |a| = 1 \) and \( |c| = 3 \), we have:
\[ |a \times c| = 1 \cdot 3 \cdot \sin(\phi) = 3 \sin(\phi) \]

### Step 4: Evaluate \( |(a \times c) \cdot d|^2 \)
Since \( d \) is a unit vector coplanar with \( b \) and \( c \), we can express:
\[ |(a \times c) \cdot d| = |a \times c| \cdot |d| \cdot \cos(\theta') \]
where \( \theta' \) is the angle between \( a \times c \) and \( d \). Since \( |d| = 1 \), we have:
\[ |(a \times c) \cdot d| = |a \times c| \cdot \cos(\theta') \]

### Step 5: Evaluate \( |(a \times c) \times \hat{d}|^2 \)
Using the formula for the magnitude of a cross product:
\[ |(a \times c) \times \hat{d}| = |a \times c| \cdot |\hat{d}| \cdot \sin(\theta'') \]
where \( \theta'' \) is the angle between \( a \times c \) and \( \hat{d} \). Since \( |\hat{d}| = 1 \), we have:
\[ |(a \times c) \times \hat{d}| = |a \times c| \cdot \sin(\theta'') \]

### Step 6: Combine the results
Now we can combine the results:
\[ |(a \times c) \cdot d|^2 + |(a \times c) \times \hat{d}|^2 = (|a \times c| \cdot \cos(\theta'))^2 + (|a \times c| \cdot \sin(\theta''))^2 \]
Using the identity \( \cos^2(\theta) + \sin^2(\theta) = 1 \):
\[ = |a \times c|^2 \cdot (\cos^2(\theta') + \sin^2(\theta'')) = |a \times c|^2 \]

### Step 7: Substitute the values
We know:
\[ |a \times c| = 3 \]
Thus:
\[ |(a \times c)|^2 = 3^2 = 9 \]

### Final Result
Therefore, the final answer is:
\[ |(a \times c) \cdot d|^2 + |(a \times c) \times \hat{d}|^2 = 9 \]