`vecA=(2veci+veck),vecB=(veci+vecj+veck) and vecC=4veci-vec3j+7veck` determine a `vecR` satisfying `vecRxxvecB=vecCxxvecB and vecR.vecA=0`

Explanation

To solve the problem, we need to determine the vector \(\vec{R}\) that satisfies two conditions:

1. \(\vec{R} \times \vec{B} = \vec{C} \times \vec{B}\)
2. \(\vec{R} \cdot \vec{A} = 0\)

Given:
- \(\vec{A} = 2\vec{i} + \vec{k}\)
- \(\vec{B} = \vec{i} + \vec{j} + \vec{k}\)
- \(\vec{C} = 4\vec{i} - 3\vec{j} + 7\vec{k}\)

### Step 1: Calculate \(\vec{C} \times \vec{B}\)

Using the determinant method for the cross product:

\[ \vec{C} \times \vec{B} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 4 & -3 & 7 \\ 1 & 1 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{C} \times \vec{B} = \vec{i} \begin{vmatrix} -3 & 7 \\ 1 & 1 \end{vmatrix} - \vec{j} \begin{vmatrix} 4 & 7 \\ 1 & 1 \end{vmatrix} + \vec{k} \begin{vmatrix} 4 & -3 \\ 1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} -3 & 7 \\ 1 & 1 \end{vmatrix} = (-3)(1) - (7)(1) = -3 - 7 = -10\)

2. \(\begin{vmatrix} 4 & 7 \\ 1 & 1 \end{vmatrix} = (4)(1) - (7)(1) = 4 - 7 = -3\)

3. \(\begin{vmatrix} 4 & -3 \\ 1 & 1 \end{vmatrix} = (4)(1) - (-3)(1) = 4 + 3 = 7\)

Putting it all together:

\[ \vec{C} \times \vec{B} = -10\vec{i} + 3\vec{j} + 7\vec{k} \]

### Step 2: Set up the equation for \(\vec{R}\)

From the first condition, we have:

\[ \vec{R} \times \vec{B} = \vec{C} \times \vec{B} \]

This implies:

\[ \vec{R} - \vec{C} \text{ is parallel to } \vec{B} \]

Thus, we can express \(\vec{R}\) as:

\[ \vec{R} = \vec{C} + t\vec{B} \]

for some scalar \(t\).

### Step 3: Substitute \(\vec{C}\) and \(\vec{B}\)

Substituting the values of \(\vec{C}\) and \(\vec{B}\):

\[ \vec{R} = (4\vec{i} - 3\vec{j} + 7\vec{k}) + t(\vec{i} + \vec{j} + \vec{k}) \]

This expands to:

\[ \vec{R} = (4 + t)\vec{i} + (-3 + t)\vec{j} + (7 + t)\vec{k} \]

### Step 4: Use the second condition \(\vec{R} \cdot \vec{A} = 0\)

Now, we need to satisfy the second condition:

\[ \vec{R} \cdot \vec{A} = 0 \]

Calculating the dot product:

\[ \vec{R} \cdot \vec{A} = (4 + t)(2) + (-3 + t)(0) + (7 + t)(1) = 0 \]

This simplifies to:

\[ 2(4 + t) + (7 + t) = 0 \]

Expanding this gives:

\[ 8 + 2t + 7 + t = 0 \]

Combining like terms:

\[ 15 + 3t = 0 \]

### Step 5: Solve for \(t\)

Solving for \(t\):

\[ 3t = -15 \implies t = -5 \]

### Step 6: Substitute \(t\) back into \(\vec{R}\)

Now substitute \(t = -5\) back into the equation for \(\vec{R}\):

\[ \vec{R} = (4 - 5)\vec{i} + (-3 - 5)\vec{j} + (7 - 5)\vec{k} \]

This simplifies to:

\[ \vec{R} = -\vec{i} - 8\vec{j} + 2\vec{k} \]

### Final Result

Thus, the vector \(\vec{R}\) is:

\[ \vec{R} = -\vec{i} - 8\vec{j} + 2\vec{k} \]