The value of `'lambda'` such that the vectors :
`3hat(i)+lambdahat(j)+5hat(k), hat(i)+2hat(j)-3hat(k)` and `2hat(i)-hat(j)+hat(k)` are coplanar is __________.

Answer

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

### Step 1: Write the vectors in component form.
The vectors are:
- \( \mathbf{A} = (3, \lambda, 5) \)
- \( \mathbf{B} = (1, 2, -3) \)
- \( \mathbf{C} = (2, -1, 1) \)

### Step 2: Set up the determinant for the scalar triple product.
The scalar triple product can be represented as the determinant of a matrix formed by the components of the vectors:
\[ \begin{vmatrix} 3 & \lambda & 5 \\ 1 & 2 & -3 \\ 2 & -1 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant.
Using the determinant formula for a 3x3 matrix:
\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where:
- \( a = 3, b = \lambda, c = 5 \)
- \( d = 1, e = 2, f = -3 \)
- \( g = 2, h = -1, i = 1 \)

The determinant becomes:
\[ \text{Det} = 3 \begin{vmatrix} 2 & -3 \\ -1 & 1 \end{vmatrix} - \lambda \begin{vmatrix} 1 & -3 \\ 2 & 1 \end{vmatrix} + 5 \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 2 & -3 \\ -1 & 1 \end{vmatrix} = (2)(1) - (-3)(-1) = 2 - 3 = -1 \)

2. \( \begin{vmatrix} 1 & -3 \\ 2 & 1 \end{vmatrix} = (1)(1) - (-3)(2) = 1 + 6 = 7 \)

3. \( \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (2)(2) = -1 - 4 = -5 \)

### Step 4: Substitute back into the determinant equation.
Now substituting these values back, we have:
\[ \text{Det} = 3(-1) - \lambda(7) + 5(-5) \]
\[ = -3 - 7\lambda - 25 \]
\[ = -28 - 7\lambda \]

### Step 5: Set the determinant to zero for coplanarity.
For the vectors to be coplanar, we set the determinant to zero:
\[ -28 - 7\lambda = 0 \]

### Step 6: Solve for \( \lambda \).
\[ -7\lambda = 28 \]
\[ \lambda = -\frac{28}{7} = -4 \]

### Final Answer:
The value of \( \lambda \) is \( -4 \).

---