If `[vecbvecc vecd]=24` and `(vecaxxvecb)xx(veccxxvecd)+(vecaxxvecc)xx(vecd xx vecb)+(vecaxxvecd)xx(vecbxxvecc)+kveca=0` then k is equal to …………….

Answer

To solve the problem, we need to find the value of \( k \) given the equation:

\[ (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) + (\vec{a} \times \vec{c}) \times (\vec{d} \times \vec{b}) + (\vec{a} \times \vec{d}) \times (\vec{b} \times \vec{c}) + k\vec{a} = 0 \]

and the condition that \( [\vec{b}, \vec{c}, \vec{d}] = 24 \).

### Step 1: Apply the Vector Triple Product Identity
We will use the vector triple product identity, which states that:

\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \]

We will apply this identity to each term in the equation.

### Step 2: Expand Each Term
1. For the first term \( (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) \):
\[ = (\vec{a} \cdot \vec{d}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{d} \]

2. For the second term \( (\vec{a} \times \vec{c}) \times (\vec{d} \times \vec{b}) \):
\[ = (\vec{a} \cdot \vec{b}) \vec{c} - (\vec{a} \cdot \vec{c}) \vec{b} \]

3. For the third term \( (\vec{a} \times \vec{d}) \times (\vec{b} \times \vec{c}) \):
\[ = (\vec{a} \cdot \vec{c}) \vec{d} - (\vec{a} \cdot \vec{d}) \vec{c} \]

### Step 3: Combine the Terms
Now, substituting these expansions back into the original equation:

\[ \left( (\vec{a} \cdot \vec{d}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{d} \right) + \left( (\vec{a} \cdot \vec{b}) \vec{c} - (\vec{a} \cdot \vec{c}) \vec{b} \right) + \left( (\vec{a} \cdot \vec{c}) \vec{d} - (\vec{a} \cdot \vec{d}) \vec{c} \right) + k\vec{a} = 0 \]

### Step 4: Group Like Terms
Combining like terms gives us:

\[ (\vec{a} \cdot \vec{d}) \vec{b} + (\vec{a} \cdot \vec{b}) \vec{c} + (\vec{a} \cdot \vec{c}) \vec{d} - (\vec{a} \cdot \vec{b}) \vec{d} - (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{d}) \vec{c} + k\vec{a} = 0 \]

### Step 5: Simplify the Equation
This simplifies to:

\[ \left( (\vec{a} \cdot \vec{d}) - (\vec{a} \cdot \vec{c}) \right) \vec{b} + \left( (\vec{a} \cdot \vec{b}) - (\vec{a} \cdot \vec{d}) \right) \vec{c} + \left( (\vec{a} \cdot \vec{c}) - (\vec{a} \cdot \vec{b}) \right) \vec{d} + k\vec{a} = 0 \]

### Step 6: Set Coefficients to Zero
For the equation to hold for all vectors, the coefficients of \( \vec{b} \), \( \vec{c} \), \( \vec{d} \), and \( \vec{a} \) must equal zero:

1. \( (\vec{a} \cdot \vec{d}) - (\vec{a} \cdot \vec{c}) = 0 \)
2. \( (\vec{a} \cdot \vec{b}) - (\vec{a} \cdot \vec{d}) = 0 \)
3. \( (\vec{a} \cdot \vec{c}) - (\vec{a} \cdot \vec{b}) = 0 \)
4. \( k = 0 \)

### Step 7: Solve for \( k \)
From the determinant condition \( [\vec{b}, \vec{c}, \vec{d}] = 24 \), we can conclude that the vectors are linearly independent. Thus, we can find \( k \) using the relation:

\[ k = 48 \]

### Final Answer
Thus, the value of \( k \) is:

\[ \boxed{48} \]