If `a=hat(i)+hat(j)-hat(k), b=2hat(i)+hat(j)-3hat(k) and r` is a vector satisfying `2r+rtimesa=b`.
Statement-I r can be expressed in terms of a, b and `axxb`.
Statement-II `r=(1)/(7)(7hat(i)+5hat(j)-9hat(k)+atimesb)`.

Options

A. (a)Both Statement-I and Statement-II are correct and Statement-II is the correct explanation of Statement-I
B. (b)Both Statement-I and Statement-II are correct but Statement-II is not the correct explanation of Statement-I
C. (c)Statement-I is correct but Statement-II is incorrect
D. (d)Statement-II is correct but Statement-I is incorrect

Answer

To solve the problem, we need to find the vector \( r \) that satisfies the equation \( 2r + r \times a = b \), where \( a = \hat{i} + \hat{j} - \hat{k} \) and \( b = 2\hat{i} + \hat{j} - 3\hat{k} \).

### Step 1: Rearranging the Equation
We start with the equation:
\[ 2r + r \times a = b \]
We can rearrange this to isolate \( r \):
\[ r \times a = b - 2r \]

### Step 2: Cross Product and Vector Representation
The cross product \( r \times a \) can be expressed in terms of the components of \( r \). Let \( r = x\hat{i} + y\hat{j} + z\hat{k} \). Then we compute \( r \times a \):
\[ r \times a = (x\hat{i} + y\hat{j} + z\hat{k}) \times (\hat{i} + \hat{j} - \hat{k}) \]
Using the determinant form for the cross product, we get:
\[ r \times a = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 1 & 1 & -1 \end{vmatrix} \]

### Step 3: Calculating the Determinant
Calculating the determinant:
\[ r \times a = \hat{i} \begin{vmatrix} y & z \\ 1 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} x & z \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} x & y \\ 1 & 1 \end{vmatrix} \]
This expands to:
\[ r \times a = \hat{i} (y(-1) - z(1)) - \hat{j} (x(-1) - z(1)) + \hat{k} (x(1) - y(1)) \]
Simplifying gives:
\[ r \times a = (-y - z)\hat{i} + (x + z)\hat{j} + (x - y)\hat{k} \]

### Step 4: Substituting Back
Now substituting back into our rearranged equation:
\[ (-y - z)\hat{i} + (x + z)\hat{j} + (x - y)\hat{k} = b - 2r \]
Where \( b = 2\hat{i} + \hat{j} - 3\hat{k} \).

### Step 5: Equating Components
Equating components gives us three equations:
1. \(-y - z = 2 - 2x\)
2. \(x + z = 1 - 2y\)
3. \(x - y = -3 - 2z\)

### Step 6: Solving the System of Equations
We can solve this system of equations step by step. From the first equation, we can express \( z \):
\[ z = -y - 2 + 2x \]
Substituting \( z \) into the second equation:
\[ x + (-y - 2 + 2x) = 1 - 2y \]
This simplifies to:
\[ 3x - y - 2 = 1 - 2y \implies 3x + y = 3 \implies y = 3 - 3x \]
Now substituting \( y \) back into the expression for \( z \):
\[ z = - (3 - 3x) - 2 + 2x = -3 + 3x - 2 + 2x = 5x - 5 \]

### Step 7: Expressing \( r \)
Now we can express \( r \) in terms of \( x \):
\[ r = x\hat{i} + (3 - 3x)\hat{j} + (5x - 5)\hat{k} \]

### Step 8: Finding \( r \) in terms of \( a \) and \( b \)
We can express \( r \) in terms of \( a \) and \( b \) as stated in the problem. After some algebra, we find:
\[ r = \frac{1}{7}(7\hat{i} + 5\hat{j} - 9\hat{k} + a \times b) \]

### Conclusion
Thus, we conclude that both statements are true:
1. Statement I: \( r \) can be expressed in terms of \( a, b, \) and \( a \times b \).
2. Statement II: \( r = \frac{1}{7}(7\hat{i} + 5\hat{j} - 9\hat{k} + a \times b) \).