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

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 is zero. This can be expressed using the determinant of a matrix formed by the components of the vectors.

### Step-by-step Solution:

1. **Write the vectors in component form:**
\[ \mathbf{a} = \begin{pmatrix} 3 \\ \lambda \\ 5 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}, \quad \mathbf{c} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \]

2. **Set up the determinant:**
The vectors are coplanar if the determinant of the matrix formed by these vectors is zero:
\[ \begin{vmatrix} 3 & \lambda & 5 \\ 1 & 2 & -3 \\ 2 & -1 & 1 \end{vmatrix} = 0 \]

3. **Calculate the determinant:**
Using the determinant formula for a 3x3 matrix:
\[ \text{Det} = 3 \begin{vmatrix} 2 & -3 \\ -1 & 1 \end{vmatrix} - \lambda \begin{vmatrix} 1 & -3 \\ 2 & 1 \end{vmatrix} + 5 \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} \]

Now calculate each of the 2x2 determinants:
- \( \begin{vmatrix} 2 & -3 \\ -1 & 1 \end{vmatrix} = (2)(1) - (-3)(-1) = 2 - 3 = -1 \)
- \( \begin{vmatrix} 1 & -3 \\ 2 & 1 \end{vmatrix} = (1)(1) - (-3)(2) = 1 + 6 = 7 \)
- \( \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (2)(2) = -1 - 4 = -5 \)

Substitute these values back into the determinant:
\[ \text{Det} = 3(-1) - \lambda(7) + 5(-5) \]
\[ \text{Det} = -3 - 7\lambda - 25 \]
\[ \text{Det} = -28 - 7\lambda \]

4. **Set the determinant to zero:**
\[ -28 - 7\lambda = 0 \]

5. **Solve for \( \lambda \):**
\[ -7\lambda = 28 \]
\[ \lambda = -4 \]

### Final Answer:
The value of \( \lambda \) such that the vectors are coplanar is \( \lambda = -4 \).