If `vecd=lambda(vecaxxvecb)+mu(vecbxxvecc)+t(veccxxveca).[veca,vecbvecc]=(1)/(8)` and `vecd.(veca+vecb+vecc)=8` then `lambda+mu+t` equals …………

Answer

To solve the problem, we need to find the value of \( \lambda + \mu + t \) given the equations involving vectors and their scalar triple products. Let's break down the solution step by step.

### Step 1: Write down the expression for vector \( \vec{d} \)
We are given:
\[ \vec{d} = \lambda (\vec{a} \times \vec{b}) + \mu (\vec{b} \times \vec{c}) + t (\vec{c} \times \vec{a}) \]

### Step 2: Take the dot product of \( \vec{d} \) with \( \vec{a} \)
Taking the dot product of both sides with \( \vec{a} \):
\[ \vec{d} \cdot \vec{a} = \lambda (\vec{a} \times \vec{b}) \cdot \vec{a} + \mu (\vec{b} \times \vec{c}) \cdot \vec{a} + t (\vec{c} \times \vec{a}) \cdot \vec{a} \]
Since \( \vec{a} \times \vec{b} \) and \( \vec{c} \times \vec{a} \) are perpendicular to \( \vec{a} \), their dot products with \( \vec{a} \) will be zero:
\[ \vec{d} \cdot \vec{a} = 0 + \mu (\vec{b} \times \vec{c}) \cdot \vec{a} + 0 \]
Using the scalar triple product:
\[ \vec{d} \cdot \vec{a} = \mu \cdot (\vec{a} \cdot (\vec{b} \times \vec{c})) = \mu \cdot 8 \]
Thus,
\[ \vec{d} \cdot \vec{a} = \frac{\mu}{8} \]

### Step 3: Take the dot product of \( \vec{d} \) with \( \vec{b} \)
Now, we take the dot product of \( \vec{d} \) with \( \vec{b} \):
\[ \vec{d} \cdot \vec{b} = \lambda (\vec{a} \times \vec{b}) \cdot \vec{b} + \mu (\vec{b} \times \vec{c}) \cdot \vec{b} + t (\vec{c} \times \vec{a}) \cdot \vec{b} \]
Again, \( \vec{a} \times \vec{b} \) and \( \vec{b} \times \vec{c} \) are perpendicular to \( \vec{b} \):
\[ \vec{d} \cdot \vec{b} = 0 + 0 + t (\vec{c} \times \vec{a}) \cdot \vec{b} \]
Using the scalar triple product:
\[ \vec{d} \cdot \vec{b} = t \cdot (\vec{b} \cdot (\vec{c} \times \vec{a})) = t \cdot 8 \]
Thus,
\[ \vec{d} \cdot \vec{b} = \frac{t}{8} \]

### Step 4: Take the dot product of \( \vec{d} \) with \( \vec{c} \)
Next, we take the dot product of \( \vec{d} \) with \( \vec{c} \):
\[ \vec{d} \cdot \vec{c} = \lambda (\vec{a} \times \vec{b}) \cdot \vec{c} + \mu (\vec{b} \times \vec{c}) \cdot \vec{c} + t (\vec{c} \times \vec{a}) \cdot \vec{c} \]
Again, \( \vec{b} \times \vec{c} \) and \( \vec{c} \times \vec{a} \) are perpendicular to \( \vec{c} \):
\[ \vec{d} \cdot \vec{c} = \lambda (\vec{a} \times \vec{b}) \cdot \vec{c} + 0 + 0 \]
Using the scalar triple product:
\[ \vec{d} \cdot \vec{c} = \lambda \cdot 8 \]
Thus,
\[ \vec{d} \cdot \vec{c} = \frac{\lambda}{8} \]

### Step 5: Combine the results
Now, we have:
\[ \vec{d} \cdot \vec{a} = \frac{\mu}{8}, \quad \vec{d} \cdot \vec{b} = \frac{t}{8}, \quad \vec{d} \cdot \vec{c} = \frac{\lambda}{8} \]
Adding these results:
\[ \vec{d} \cdot \vec{a} + \vec{d} \cdot \vec{b} + \vec{d} \cdot \vec{c} = \frac{\mu + t + \lambda}{8} \]

### Step 6: Use the given condition
We are given that:
\[ \vec{d} \cdot (\vec{a} + \vec{b} + \vec{c}) = 8 \]
Thus, we can equate:
\[ \frac{\mu + t + \lambda}{8} = 8 \]
Multiplying both sides by 8:
\[ \mu + t + \lambda = 64 \]

### Final Answer
Thus, the value of \( \lambda + \mu + t \) is:
\[ \boxed{64} \]