Find the value of 'x' such that four points with position vectors : `A(3hat(i)-2hat(j)+hat(k)), B(4hat(i)+x hat(j)+5hat(k)), C(4hat(i)+2hat(j)-2hat(k))` and `D(6hat(i)+5hat(j)-hat(k))` are coplanar.

Answer

To find the value of 'x' such that the points A, B, C, and D with given position vectors are coplanar, we can use the scalar triple product condition. The points are coplanar if the scalar triple product of the vectors AB, AC, and AD is zero.

### Step-by-Step Solution:

1. **Identify the Position Vectors**:
- \( \vec{A} = 3\hat{i} - 2\hat{j} + \hat{k} \)
- \( \vec{B} = 4\hat{i} + x\hat{j} + 5\hat{k} \)
- \( \vec{C} = 4\hat{i} + 2\hat{j} - 2\hat{k} \)
- \( \vec{D} = 6\hat{i} + 5\hat{j} - \hat{k} \)

2. **Find the Vectors AB, AC, and AD**:
- **Vector AB**:
\[ \vec{AB} = \vec{B} - \vec{A} = (4\hat{i} + x\hat{j} + 5\hat{k}) - (3\hat{i} - 2\hat{j} + \hat{k}) = (4 - 3)\hat{i} + (x + 2)\hat{j} + (5 - 1)\hat{k} = \hat{i} + (x + 2)\hat{j} + 4\hat{k} \]
- **Vector AC**:
\[ \vec{AC} = \vec{C} - \vec{A} = (4\hat{i} + 2\hat{j} - 2\hat{k}) - (3\hat{i} - 2\hat{j} + \hat{k}) = (4 - 3)\hat{i} + (2 + 2)\hat{j} + (-2 - 1)\hat{k} = \hat{i} + 4\hat{j} - 3\hat{k} \]
- **Vector AD**:
\[ \vec{AD} = \vec{D} - \vec{A} = (6\hat{i} + 5\hat{j} - \hat{k}) - (3\hat{i} - 2\hat{j} + \hat{k}) = (6 - 3)\hat{i} + (5 + 2)\hat{j} + (-1 - 1)\hat{k} = 3\hat{i} + 7\hat{j} - 2\hat{k} \]

3. **Set Up the Scalar Triple Product**:
The scalar triple product can be calculated using the determinant:
\[ \left| \begin{array}{ccc} 1 & x + 2 & 4 \\ 1 & 4 & -3 \\ 3 & 7 & -2 \end{array} \right| = 0 \]

4. **Calculate the Determinant**:
Expanding the determinant:
\[ = 1 \left| \begin{array}{cc} 4 & -3 \\ 7 & -2 \end{array} \right| - (x + 2) \left| \begin{array}{cc} 1 & -3 \\ 3 & -2 \end{array} \right| + 4 \left| \begin{array}{cc} 1 & 4 \\ 3 & 7 \end{array} \right| \]
Calculating each of these 2x2 determinants:
\[ = 1(4 \cdot -2 - (-3) \cdot 7) - (x + 2)(1 \cdot -2 - (-3) \cdot 3) + 4(1 \cdot 7 - 4 \cdot 3) \]
\[ = 1(-8 + 21) - (x + 2)(-2 + 9) + 4(7 - 12) \]
\[ = 1(13) - (x + 2)(7) + 4(-5) \]
\[ = 13 - 7(x + 2) - 20 \]
\[ = 13 - 7x - 14 - 20 = -7x - 21 \]

5. **Set the Determinant to Zero**:
\[ -7x - 21 = 0 \]
\[ -7x = 21 \]
\[ x = -3 \]

### Final Answer:
The value of \( x \) such that points A, B, C, and D are coplanar is \( x = -3 \).