If `a = (-1,1,1) and b = (2,0,1)` then the vector r satisfying the conditions
(i) that it is coplanar with a and b
(ii) that it is perpendicular to b
(iii) that `a.r = 7` is

Options

A. `-3i+4j+6k`
B. `-(3)/(2)i+(5)/(2)j+3k`
C. `3i+16j-6k`
D. none of these

Answer

To find the vector \( \mathbf{r} \) that satisfies the given conditions, we will follow a systematic approach.

### Given:
- \( \mathbf{a} = (-1, 1, 1) \)
- \( \mathbf{b} = (2, 0, 1) \)

### Conditions:
1. \( \mathbf{r} \) is coplanar with \( \mathbf{a} \) and \( \mathbf{b} \).
2. \( \mathbf{r} \) is perpendicular to \( \mathbf{b} \).
3. \( \mathbf{a} \cdot \mathbf{r} = 7 \).

### Step 1: Express \( \mathbf{r} \) in terms of \( \mathbf{a} \) and \( \mathbf{b} \)

Since \( \mathbf{r} \) is coplanar with \( \mathbf{a} \) and \( \mathbf{b} \), we can write:
\[ \mathbf{r} = \lambda \mathbf{a} + \mu \mathbf{b} \]
where \( \lambda \) and \( \mu \) are scalars.

Substituting the vectors:
\[ \mathbf{r} = \lambda (-1, 1, 1) + \mu (2, 0, 1) = (-\lambda + 2\mu, \lambda, \lambda + \mu) \]

### Step 2: Use the condition that \( \mathbf{r} \) is perpendicular to \( \mathbf{b} \)

For \( \mathbf{r} \) to be perpendicular to \( \mathbf{b} \), their dot product must equal zero:
\[ \mathbf{r} \cdot \mathbf{b} = 0 \]
Calculating the dot product:
\[ (-\lambda + 2\mu, \lambda, \lambda + \mu) \cdot (2, 0, 1) = 2(-\lambda + 2\mu) + 0 \cdot \lambda + 1(\lambda + \mu) = 0 \]
This simplifies to:
\[ -2\lambda + 4\mu + \lambda + \mu = 0 \]
Combining like terms:
\[ -2\lambda + \lambda + 5\mu = 0 \implies -\lambda + 5\mu = 0 \implies \lambda = 5\mu \]

### Step 3: Use the condition \( \mathbf{a} \cdot \mathbf{r} = 7 \)

Now we substitute \( \lambda = 5\mu \) into the dot product condition:
\[ \mathbf{a} \cdot \mathbf{r} = 7 \]
Calculating the dot product:
\[ (-1, 1, 1) \cdot (-\lambda + 2\mu, \lambda, \lambda + \mu) = -(-\lambda + 2\mu) + \lambda + (\lambda + \mu) = 7 \]
Substituting \( \lambda = 5\mu \):
\[ -(-5\mu + 2\mu) + 5\mu + (5\mu + \mu) = 7 \]
This simplifies to:
\[ (5\mu - 2\mu) + 5\mu + 6\mu = 7 \implies 5\mu + 5\mu + 6\mu = 7 \implies 16\mu = 7 \implies \mu = \frac{7}{16} \]

### Step 4: Find \( \lambda \)

Now substituting back to find \( \lambda \):
\[ \lambda = 5\mu = 5 \cdot \frac{7}{16} = \frac{35}{16} \]

### Step 5: Substitute \( \lambda \) and \( \mu \) back into \( \mathbf{r} \)

Now substituting \( \lambda \) and \( \mu \) into the expression for \( \mathbf{r} \):
\[ \mathbf{r} = \left(-\frac{35}{16} + 2 \cdot \frac{7}{16}, \frac{35}{16}, \frac{35}{16} + \frac{7}{16}\right) \]
Calculating each component:
1. First component:
\[ -\frac{35}{16} + \frac{14}{16} = -\frac{21}{16} \]
2. Second component:
\[ \frac{35}{16} \]
3. Third component:
\[ \frac{35}{16} + \frac{7}{16} = \frac{42}{16} = \frac{21}{8} \]

Thus, the vector \( \mathbf{r} \) is:
\[ \mathbf{r} = \left(-\frac{21}{16}, \frac{35}{16}, \frac{21}{8}\right) \]

### Final Answer:
\[ \mathbf{r} = \left(-\frac{21}{16}, \frac{35}{16}, \frac{21}{8}\right) \]