Consider the three vectors p, q, r such that `p=hat(i)+hat(j)+hat(k) and q=hat(i)-hat(j)+hat(k) , ptimesr=q+cp and p*r=2`
Q.The value of [p q r] is

Options

A. `(5sqrt(2)c)/(|r|)`
B. `-(8)/(3)`
C. `0`
D. greater than 0

Answer

To solve the problem, we need to find the scalar triple product \([p, q, r]\) given the vectors \(p\), \(q\), and \(r\) along with some conditions. Let's break down the solution step by step.

### Given:
- \( p = \hat{i} + \hat{j} + \hat{k} \)
- \( q = \hat{i} - \hat{j} + \hat{k} \)
- \( p \times r = q + cp \)
- \( p \cdot r = 2 \)

### Step 1: Define the vector \(r\)
Let \( r = x\hat{i} + y\hat{j} + z\hat{k} \).

### Step 2: Calculate \( p \times r \)
Using the determinant method for the cross product:

\[ p \times r = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} \]

Calculating the determinant:

\[ p \times r = \hat{i}(1 \cdot z - 1 \cdot y) - \hat{j}(1 \cdot z - 1 \cdot x) + \hat{k}(1 \cdot y - 1 \cdot x) \]

This simplifies to:

\[ p \times r = (z - y)\hat{i} - (z - x)\hat{j} + (y - x)\hat{k} \]

### Step 3: Set up the equation from \( p \times r = q + cp \)
We know:

\[ q + cp = (\hat{i} - \hat{j} + \hat{k}) + c(\hat{i} + \hat{j} + \hat{k}) = (1+c)\hat{i} + (-1+c)\hat{j} + (1+c)\hat{k} \]

Setting the two expressions equal gives us:

\[ (z - y)\hat{i} - (z - x)\hat{j} + (y - x)\hat{k} = (1+c)\hat{i} + (-1+c)\hat{j} + (1+c)\hat{k} \]

### Step 4: Equate coefficients
From the above equation, we can equate coefficients:

1. \( z - y = 1 + c \) (Equation 1)
2. \( -(z - x) = -1 + c \) (Equation 2) → \( z - x = 1 - c \) (Equation 2)
3. \( y - x = 1 + c \) (Equation 3)

### Step 5: Solve the system of equations
From Equations 1 and 2:

From Equation 1: \( z = y + 1 + c \)

Substituting \(z\) in Equation 2:

\[ y + 1 + c - x = 1 - c \implies y - x + 2c = 0 \implies x = y + 2c \quad (Equation 4) \]

Now substituting \(x\) in Equation 3:

\[ y - (y + 2c) = 1 + c \implies -2c = 1 + c \implies -3c = 1 \implies c = -\frac{1}{3} \]

### Step 6: Substitute \(c\) back to find \(x\), \(y\), and \(z\)
Using \(c = -\frac{1}{3}\):

From Equation 4:

\[ x = y - \frac{2}{3} \]

Substituting \(c\) into Equation 1:

\[ z = y + 1 - \frac{1}{3} = y + \frac{2}{3} \]

### Step 7: Use the dot product condition
Using \(p \cdot r = 2\):

\[ 1 + y + z = 2 \implies 1 + y + (y + \frac{2}{3}) = 2 \implies 2y + \frac{5}{3} = 2 \implies 2y = \frac{1}{3} \implies y = \frac{1}{6} \]

Now substituting \(y\) back to find \(x\) and \(z\):

\[ x = \frac{1}{6} - \frac{2}{3} = \frac{1}{6} - \frac{4}{6} = -\frac{1}{2} \]
\[ z = \frac{1}{6} + \frac{2}{3} = \frac{1}{6} + \frac{4}{6} = \frac{5}{6} \]

### Step 8: Construct vector \(r\)
Thus, we have:

\[ r = -\frac{1}{2}\hat{i} + \frac{1}{6}\hat{j} + \frac{5}{6}\hat{k} \]

### Step 9: Calculate the scalar triple product \([p, q, r]\)
Using the determinant method:

\[ [p, q, r] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ -\frac{1}{2} & \frac{1}{6} & \frac{5}{6} \end{vmatrix} \]

Calculating the determinant:

\[ = 1 \left( -1 \cdot \frac{5}{6} - 1 \cdot \frac{1}{6} \right) - 1 \left( 1 \cdot \frac{5}{6} - 1 \cdot -\frac{1}{2} \right) + 1 \left( 1 \cdot \frac{1}{6} - (-1) \cdot -\frac{1}{2} \right) \]

Calculating each term:

1. First term: \( -1 \cdot \frac{6}{6} = -1 \)
2. Second term: \( 1 \cdot \left( \frac{5}{6} + \frac{3}{6} \right) = 1 \cdot \frac{8}{6} = \frac{4}{3} \)
3. Third term: \( 1 \cdot \left( \frac{1}{6} - \frac{1}{2} \right) = 1 \cdot \left( \frac{1}{6} - \frac{3}{6} \right) = 1 \cdot -\frac{2}{6} = -\frac{1}{3} \)

Putting it all together:

\[ = -1 - \frac{4}{3} - \frac{1}{3} = -1 - \frac{5}{3} = -\frac{8}{3} \]

### Final Answer:
Thus, the value of the scalar triple product \([p, q, r]\) is:

\[ \boxed{-\frac{8}{3}} \]