To solve the given vector expression, let's break it down step by step. We need to find the value of:

\((\vec{r} \cdot \hat{i})(\vec{r} \times \hat{i}) + (\vec{r} \cdot \hat{j})(\vec{r} \times \hat{j}) + (\vec{r} \cdot \hat{k})(\vec{r} \times \hat{k})\)

where \(\vec{r}\) is any vector in 3D space, and \(\hat{i}, \hat{j}, \hat{k}\) are the unit vectors along the x, y, and z axes, respectively.

1. Consider the term \((\vec{r} \cdot \hat{i})(\vec{r} \times \hat{i})\). Here,:
   • \(\vec{r} \cdot \hat{i}\) represents the component of \(\vec{r}\) along the x-axis, which is \(x\), if \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\).
   • \(\vec{r} \times \hat{i}\) is the cross product of \(\vec{r}\) and \(\hat{i}\). For any vector cross product with itself or its unit vector, the result will include only the orthogonal components. Thus, we get:
     • \(\vec{r} \times \hat{i} = (x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{i} = y(\hat{j} \times \hat{i}) + z(\hat{k} \times \hat{i}) = -y\hat{k} + z\hat{j}\)
   • Since \(\vec{r} \cdot \hat{i} = x\), we have \((\vec{r} \cdot \hat{i})(\vec{r} \times \hat{i}) = x(-y\hat{k} + z\hat{j}) = -xy\hat{k} + xz\hat{j}\).
2. The same logic applies to the remaining terms:
   • \((\vec{r} \cdot \hat{j})(\vec{r} \times \hat{j})\):
     • \(\vec{r} \cdot \hat{j} = y\)
     • \(\vec{r} \times \hat{j} = (z\hat{i} - x\hat{k})\)
     • Thus, \(y(z\hat{i} - x\hat{k}) = yz\hat{i} - yx\hat{k}\)
   • \((\vec{r} \cdot \hat{k})(\vec{r} \times \hat{k})\):
     • \(\vec{r} \cdot \hat{k} = z\)
     • \(\vec{r} \times \hat{k} = (x\hat{j} - y\hat{i})\)
     • Thus, \(z(x\hat{j} - y\hat{i}) = zx\hat{j} - zy\hat{i}\)
3. Combine all the terms:
   • Summing these vector products:
   • \(-xy\hat{k} + xz\hat{j} + yz\hat{i} - yx\hat{k} + zx\hat{j} - zy\hat{i}\) results in:
     • Terms involving \(\hat{i}\): \(yz\hat{i} - zy\hat{i} = 0\)
     • Terms involving \(\hat{j}\): \(xz\hat{j} + zx\hat{j} = 2xz\hat{j} - 2xz\hat{j} = 0\)
     • Terms involving \(\hat{k}\): \(-xy\hat{k} - yx\hat{k} = -2xy\hat{k} + 2xy\hat{k} = 0\)
   • Hence, the entire expression sums up to the zero vector: \(\vec{0}\).

Based on the detailed calculations, the correct answer is the zero vector, denoted by \(\vec{0}\).