If `bar(a)+lamdabar(b)+3bar(c),-2bar(a)+3bar(b)-4bar(c),bar(a)-3bar(b)+5bar(c)` are coplanar, then find value of `lamda`

Answer

To solve the problem, we need to determine the value of \(\lambda\) such that the vectors \(\bar{a} + \lambda \bar{b} + 3 \bar{c}\), \(-2 \bar{a} + 3 \bar{b} - 4 \bar{c}\), and \(\bar{a} - 3 \bar{b} + 5 \bar{c}\) are coplanar.

### Step-by-Step Solution:

1. **Identify the Vectors:**
Let:
\[ \bar{u} = \bar{a} + \lambda \bar{b} + 3 \bar{c} \]
\[ \bar{v} = -2 \bar{a} + 3 \bar{b} - 4 \bar{c} \]
\[ \bar{w} = \bar{a} - 3 \bar{b} + 5 \bar{c} \]

2. **Condition for Coplanarity:**
The vectors \(\bar{u}\), \(\bar{v}\), and \(\bar{w}\) are coplanar if the scalar triple product is zero:
\[ \bar{u} \cdot (\bar{v} \times \bar{w}) = 0 \]
This can be represented using the determinant of a matrix formed by the coefficients of the vectors.

3. **Set Up the Determinant:**
We can create a determinant using the coefficients of \(\bar{a}\), \(\bar{b}\), and \(\bar{c}\):
\[ \begin{vmatrix} 1 & \lambda & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix} = 0 \]

4. **Calculate the Determinant:**
Expanding the determinant:
\[ = 1 \begin{vmatrix} 3 & -4 \\ -3 & 5 \end{vmatrix} - \lambda \begin{vmatrix} -2 & -4 \\ 1 & 5 \end{vmatrix} + 3 \begin{vmatrix} -2 & 3 \\ 1 & -3 \end{vmatrix} \]

5. **Compute Each 2x2 Determinant:**
- For \( \begin{vmatrix} 3 & -4 \\ -3 & 5 \end{vmatrix} = (3)(5) - (-4)(-3) = 15 - 12 = 3 \)
- For \( \begin{vmatrix} -2 & -4 \\ 1 & 5 \end{vmatrix} = (-2)(5) - (-4)(1) = -10 + 4 = -6 \)
- For \( \begin{vmatrix} -2 & 3 \\ 1 & -3 \end{vmatrix} = (-2)(-3) - (3)(1) = 6 - 3 = 3 \)

6. **Substitute Back into the Determinant:**
\[ 1(3) - \lambda(-6) + 3(3) = 0 \]
This simplifies to:
\[ 3 + 6\lambda + 9 = 0 \]
\[ 12 + 6\lambda = 0 \]

7. **Solve for \(\lambda\):**
\[ 6\lambda = -12 \implies \lambda = -2 \]

### Final Answer:
The value of \(\lambda\) is \(-2\).