For vectors `vec(A) = (2i + 3j - 2k), vec(B) = (5i + nj + k) and vec(C ) = (-i + 2j + 3k)` to be coplanar, the value of n is

Answer

To find the value of \( n \) such that the vectors \( \vec{A} = (2\hat{i} + 3\hat{j} - 2\hat{k}) \), \( \vec{B} = (5\hat{i} + n\hat{j} + \hat{k}) \), and \( \vec{C} = (-\hat{i} + 2\hat{j} + 3\hat{k}) \) are coplanar, we will use the condition that the scalar triple product (or box product) of the three vectors must be zero.

### Step 1: Set up the determinant

The scalar triple product can be computed using the determinant of a matrix formed by the components of the vectors:

\[ \begin{vmatrix} 2 & 3 & -2 \\ 5 & n & 1 \\ -1 & 2 & 3 \end{vmatrix} \]

### Step 2: Calculate the determinant

We will expand this determinant using the first row:

\[ = 2 \begin{vmatrix} n & 1 \\ 2 & 3 \end{vmatrix} - 3 \begin{vmatrix} 5 & 1 \\ -1 & 3 \end{vmatrix} - 2 \begin{vmatrix} 5 & n \\ -1 & 2 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants

1. For the first determinant:

\[ \begin{vmatrix} n & 1 \\ 2 & 3 \end{vmatrix} = n \cdot 3 - 1 \cdot 2 = 3n - 2 \]

2. For the second determinant:

\[ \begin{vmatrix} 5 & 1 \\ -1 & 3 \end{vmatrix} = 5 \cdot 3 - 1 \cdot (-1) = 15 + 1 = 16 \]

3. For the third determinant:

\[ \begin{vmatrix} 5 & n \\ -1 & 2 \end{vmatrix} = 5 \cdot 2 - n \cdot (-1) = 10 + n \]

### Step 4: Substitute back into the determinant equation

Now substituting these back into the determinant expression:

\[ = 2(3n - 2) - 3(16) - 2(10 + n) \]

### Step 5: Simplify the expression

Expanding this gives:

\[ = 6n - 4 - 48 - 20 - 2n \]

Combining like terms:

\[ = (6n - 2n) + (-4 - 48 - 20) = 4n - 72 \]

### Step 6: Set the determinant to zero

For the vectors to be coplanar, we set the determinant to zero:

\[ 4n - 72 = 0 \]

### Step 7: Solve for \( n \)

Solving for \( n \):

\[ 4n = 72 \implies n = \frac{72}{4} = 18 \]

### Conclusion

Thus, the value of \( n \) for which the vectors are coplanar is:

\[ \boxed{18} \]

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