Consider three vectors
`vec(A)=2 hat(i)+3 hat(j)-2 hat(k)" " vec(B)=5hat(i)+nhat(j)+hat(k)" " vec(C)=-hat(i)+2hat(j)+3 hat(k)`
If these three vectors are coplanar, then value of n will be

Options

A. 0
B. 12
C. 16
D. 18

Explanation

To determine the value of \( n \) for which the vectors \( \vec{A} \), \( \vec{B} \), and \( \vec{C} \) are coplanar, we can use the condition that the scalar triple product of the vectors must equal zero. The scalar triple product can be calculated using the determinant of a matrix formed by the components of the vectors.

### Step-by-Step Solution:

1. **Write down the vectors:**
\[ \vec{A} = 2 \hat{i} + 3 \hat{j} - 2 \hat{k} \]
\[ \vec{B} = 5 \hat{i} + n \hat{j} + \hat{k} \]
\[ \vec{C} = -\hat{i} + 2 \hat{j} + 3 \hat{k} \]

2. **Set up the determinant for the scalar triple product:**
The scalar triple product can be expressed as:
\[ \vec{A} \cdot (\vec{B} \times \vec{C}) = \begin{vmatrix} 2 & 3 & -2 \\ 5 & n & 1 \\ -1 & 2 & 3 \end{vmatrix} \]

3. **Calculate the determinant:**
We can expand this determinant using the first row:
\[ = 2 \begin{vmatrix} n & 1 \\ 2 & 3 \end{vmatrix} - 3 \begin{vmatrix} 5 & 1 \\ -1 & 3 \end{vmatrix} - 2 \begin{vmatrix} 5 & n \\ -1 & 2 \end{vmatrix} \]

Now calculating each of the 2x2 determinants:
- First determinant:
\[ \begin{vmatrix} n & 1 \\ 2 & 3 \end{vmatrix} = n \cdot 3 - 1 \cdot 2 = 3n - 2 \]
- Second determinant:
\[ \begin{vmatrix} 5 & 1 \\ -1 & 3 \end{vmatrix} = 5 \cdot 3 - 1 \cdot (-1) = 15 + 1 = 16 \]
- Third determinant:
\[ \begin{vmatrix} 5 & n \\ -1 & 2 \end{vmatrix} = 5 \cdot 2 - n \cdot (-1) = 10 + n \]

4. **Substituting back into the determinant:**
\[ = 2(3n - 2) - 3(16) - 2(10 + n) \]
\[ = 6n - 4 - 48 - 20 - 2n \]
\[ = 6n - 2n - 4 - 48 - 20 \]
\[ = 4n - 72 \]

5. **Set the determinant equal to zero for coplanarity:**
\[ 4n - 72 = 0 \]

6. **Solve for \( n \):**
\[ 4n = 72 \]
\[ n = \frac{72}{4} = 18 \]

### Conclusion:
The value of \( n \) for which the vectors \( \vec{A} \), \( \vec{B} \), and \( \vec{C} \) are coplanar is:
\[ \boxed{18} \]