Step 1: Understanding the Concept:
This problem involves vector equations of lines in 3D space. The cross product equations \( (\vec{r} - \vec{b}) \times \vec{a} = 0 \) and \( (\vec{r} - \vec{a}) \times \vec{b} = 0 \) indicate that the vector \( (\vec{r} - \vec{b}) \) is collinear with \( \vec{a} \) and the vector \( (\vec{r} - \vec{a}) \) is collinear with \( \vec{b} \). This allows us to write the position vector \( \vec{r} \) in parametric form for both equations and find their unique point of intersection.
Step 2: Key Formula or Approach:
1. For \( \vec{r} \times \vec{a} = \vec{b} \times \vec{a} \), rewrite as \( (\vec{r} - \vec{b}) \times \vec{a} = 0 \). This means \( \vec{r} - \vec{b} = \lambda \vec{a} \).
2. For \( \vec{r} \times \vec{b} = \vec{a} \times \vec{b} \), rewrite as \( (\vec{r} - \vec{a}) \times \vec{b} = 0 \). This means \( \vec{r} - \vec{a} = \mu \vec{b} \).
3. Solve the system \( \vec{b} + \lambda \vec{a} = \vec{a} + \mu \vec{b} \).
Step 3: Detailed Explanation:
From the first equation:
\[ \vec{r} = \vec{b} + \lambda \vec{a} \]
From the second equation:
\[ \vec{r} = \vec{a} + \mu \vec{b} \]
Equating the two expressions for \( \vec{r} \):
\[ \vec{b} + \lambda \vec{a} = \vec{a} + \mu \vec{b} \]
\[ \lambda \vec{a} - \mu \vec{b} = \vec{a} - \vec{b} \]
\[ (\lambda - 1) \vec{a} + (1 - \mu) \vec{b} = 0 \]
Since \( \vec{a} \) and \( \vec{b} \) are non-collinear vectors (\( \hat{i} + \hat{j} \) and \( 2\hat{i} - \hat{k} \) are not multiples of each other), the only way their linear combination is zero is if the coefficients are zero:
\[ \lambda - 1 = 0 \implies \lambda = 1 \]
\[ 1 - \mu = 0 \implies \mu = 1 \]
Now, substitute \( \lambda = 1 \) back into the first equation:
\[ \vec{r} = \vec{b} + (1)\vec{a} = \vec{a} + \vec{b} \]
Given \( \vec{a} = \hat{i} + \hat{j} \) and \( \vec{b} = 2\hat{i} - \hat{k} \):
\[ \vec{r} = (\hat{i} + \hat{j}) + (2\hat{i} - \hat{k}) = 3\hat{i} + \hat{j} - \hat{k} \]
Step 4: Final Answer:
The point of intersection is \( 3\hat{i} + \hat{j} - \hat{k} \), which matches option (D).