Let `veca=hati+2hatj-hatk, vecb=hati-hatj and vecc=hati-hatj-hatk` be three given vectors. If `vecr` is a vector such that `vecr xx veca=vecc xx veca and vecr*vecb=0`, then `vecr*veca` is equal to _______.

Answer

To solve the problem, we need to find \( \vec{r} \cdot \vec{a} \) given the conditions \( \vec{r} \times \vec{a} = \vec{c} \times \vec{a} \) and \( \vec{r} \cdot \vec{b} = 0 \).

### Step-by-Step Solution:

1. **Identify the vectors**:
- \( \vec{a} = \hat{i} + 2\hat{j} - \hat{k} \)
- \( \vec{b} = \hat{i} - \hat{j} \)
- \( \vec{c} = \hat{i} - \hat{j} - \hat{k} \)

2. **Use the first condition**:
From the condition \( \vec{r} \times \vec{a} = \vec{c} \times \vec{a} \), we can rearrange it to:
\[ \vec{r} \times \vec{a} - \vec{c} \times \vec{a} = \vec{0} \]
This implies that \( \vec{r} - \vec{c} = \lambda \vec{a} \) for some scalar \( \lambda \).

3. **Express \( \vec{r} \)**:
\[ \vec{r} = \lambda \vec{a} + \vec{c} \]

4. **Use the second condition**:
The second condition is \( \vec{r} \cdot \vec{b} = 0 \). Substituting \( \vec{r} \):
\[ (\lambda \vec{a} + \vec{c}) \cdot \vec{b} = 0 \]
Expanding this gives:
\[ \lambda (\vec{a} \cdot \vec{b}) + \vec{c} \cdot \vec{b} = 0 \]

5. **Calculate \( \vec{a} \cdot \vec{b} \)**:
\[ \vec{a} \cdot \vec{b} = (1)(1) + (2)(-1) + (-1)(0) = 1 - 2 + 0 = -1 \]

6. **Calculate \( \vec{c} \cdot \vec{b} \)**:
\[ \vec{c} \cdot \vec{b} = (1)(1) + (-1)(-1) + (-1)(0) = 1 + 1 + 0 = 2 \]

7. **Substitute the dot products**:
\[ \lambda (-1) + 2 = 0 \]
Solving for \( \lambda \):
\[ -\lambda + 2 = 0 \implies \lambda = 2 \]

8. **Substitute \( \lambda \) back into \( \vec{r} \)**:
\[ \vec{r} = 2\vec{a} + \vec{c} \]
Substitute the values of \( \vec{a} \) and \( \vec{c} \):
\[ \vec{r} = 2(\hat{i} + 2\hat{j} - \hat{k}) + (\hat{i} - \hat{j} - \hat{k}) = (2\hat{i} + 4\hat{j} - 2\hat{k}) + (\hat{i} - \hat{j} - \hat{k}) \]
Simplifying this:
\[ \vec{r} = (2 + 1)\hat{i} + (4 - 1)\hat{j} + (-2 - 1)\hat{k} = 3\hat{i} + 3\hat{j} - 3\hat{k} \]

9. **Calculate \( \vec{r} \cdot \vec{a} \)**:
\[ \vec{r} \cdot \vec{a} = (3\hat{i} + 3\hat{j} - 3\hat{k}) \cdot (\hat{i} + 2\hat{j} - \hat{k}) = 3(1) + 3(2) + (-3)(-1) \]
This simplifies to:
\[ 3 + 6 + 3 = 12 \]

### Final Answer:
\[ \vec{r} \cdot \vec{a} = 12 \]