Find the value of `'lambda'` such that 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 need to set up the determinant of the coefficients of these vectors and equate it to zero.

### Step-by-Step Solution:

1. **Write the vectors in coefficient 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:
\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]
For our matrix:
\[ D = 3 \begin{vmatrix} 2 & -3 \\ -1 & 1 \end{vmatrix} - \lambda \begin{vmatrix} 1 & -3 \\ 2 & 1 \end{vmatrix} + 5 \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} \]

4. **Calculate the smaller 2x2 determinants**:
- First determinant:
\[ \begin{vmatrix} 2 & -3 \\ -1 & 1 \end{vmatrix} = (2)(1) - (-3)(-1) = 2 - 3 = -1 \]
- Second determinant:
\[ \begin{vmatrix} 1 & -3 \\ 2 & 1 \end{vmatrix} = (1)(1) - (-3)(2) = 1 + 6 = 7 \]
- Third determinant:
\[ \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (2)(2) = -1 - 4 = -5 \]

5. **Substitute back into the determinant equation**:
\[ D = 3(-1) - \lambda(7) + 5(-5) = -3 - 7\lambda - 25 \]
Simplifying gives:
\[ D = -28 - 7\lambda \]

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

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

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