If `vec(a) = hat(i) - hat(j) + hat(k). vec(b) = 2hat(i) + 3hat(j) + 2hat(k) and vec(c ) = hat(i) + m hat(j) + n hat(k)` are three coplanar vectors and `|vec(c )| = sqrt6`, then which one of the following is correct?

Options

A. `m=2 and n = +-1`
B. `m = +-2 and n = -1`
C. `m =2 and n = -1`
D. `m = +-2 and n = 1`

Answer

To solve the problem, we need to analyze the given vectors and their properties. Let's break down the solution step by step.

### Step 1: Define the vectors
We have three vectors:
- \(\vec{a} = \hat{i} - \hat{j} + \hat{k}\)
- \(\vec{b} = 2\hat{i} + 3\hat{j} + 2\hat{k}\)
- \(\vec{c} = \hat{i} + m\hat{j} + n\hat{k}\)

### Step 2: Find the modulus of vector \(\vec{c}\)
The modulus of vector \(\vec{c}\) is given as \(|\vec{c}| = \sqrt{6}\). The modulus of a vector \(\vec{c} = \hat{i} + m\hat{j} + n\hat{k}\) is calculated using the formula:
\[ |\vec{c}| = \sqrt{1^2 + m^2 + n^2} = \sqrt{1 + m^2 + n^2} \]
Setting this equal to \(\sqrt{6}\), we have:
\[ \sqrt{1 + m^2 + n^2} = \sqrt{6} \]
Squaring both sides gives:
\[ 1 + m^2 + n^2 = 6 \]
Thus, we can simplify this to:
\[ m^2 + n^2 = 5 \quad \text{(Equation 1)} \]

### Step 3: Use the coplanarity condition
The vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are coplanar. For three vectors to be coplanar, the scalar triple product (box product) must equal zero:
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = 0 \]
We can express this as a determinant:
\[ \begin{vmatrix} 1 & -1 & 1 \\ 2 & 3 & 2 \\ 1 & m & n \end{vmatrix} = 0 \]
Calculating this determinant, we expand it:
\[ = 1 \begin{vmatrix} 3 & 2 \\ m & n \end{vmatrix} - (-1) \begin{vmatrix} 2 & 2 \\ 1 & n \end{vmatrix} + 1 \begin{vmatrix} 2 & 3 \\ 1 & m \end{vmatrix} \]
Calculating each of these 2x2 determinants:
1. \(3n - 2m\)
2. \(2n - 2\)
3. \(2m - 3\)

Putting it all together:
\[ (3n - 2m) + (2n - 2) + (2m - 3) = 0 \]
Simplifying this gives:
\[ 5n + 0m - 5 = 0 \]
Thus:
\[ 5n = 5 \implies n = 1 \quad \text{(Equation 2)} \]

### Step 4: Substitute \(n\) back into Equation 1
Now that we have \(n = 1\), we can substitute this into Equation 1:
\[ m^2 + 1^2 = 5 \]
This simplifies to:
\[ m^2 + 1 = 5 \implies m^2 = 4 \]
Taking the square root gives:
\[ m = 2 \quad \text{or} \quad m = -2 \]

### Final Result
Thus, the values of \(m\) and \(n\) are:
- \(m = 2\) or \(m = -2\)
- \(n = 1\)

### Conclusion
The correct option based on the values derived is that \(m\) can be \(2\) or \(-2\) and \(n\) is \(1\).

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