What is the value of a so that the equations `ax + 2y + 3z = 0 , 2x -3 y + 4z =0` and `5x + 7y -8z =0` has unique solution

Answer

To find the value of \( a \) such that the system of equations has a unique solution, we need to ensure that the determinant of the coefficient matrix is non-zero. The equations given are:

1. \( ax + 2y + 3z = 0 \)
2. \( 2x - 3y + 4z = 0 \)
3. \( 5x + 7y - 8z = 0 \)

### Step 1: Write the coefficient matrix
The coefficient matrix \( A \) for the system of equations is:

\[ A = \begin{bmatrix} a & 2 & 3 \\ 2 & -3 & 4 \\ 5 & 7 & -8 \end{bmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix
To find the determinant of matrix \( A \), we use the formula for the determinant of a 3x3 matrix:

\[ \text{det}(A) = a \begin{vmatrix} -3 & 4 \\ 7 & -8 \end{vmatrix} - 2 \begin{vmatrix} 2 & 4 \\ 5 & -8 \end{vmatrix} + 3 \begin{vmatrix} 2 & -3 \\ 5 & 7 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} -3 & 4 \\ 7 & -8 \end{vmatrix} = (-3)(-8) - (4)(7) = 24 - 28 = -4 \)

2. \( \begin{vmatrix} 2 & 4 \\ 5 & -8 \end{vmatrix} = (2)(-8) - (4)(5) = -16 - 20 = -36 \)

3. \( \begin{vmatrix} 2 & -3 \\ 5 & 7 \end{vmatrix} = (2)(7) - (-3)(5) = 14 + 15 = 29 \)

Substituting these values back into the determinant formula:

\[ \text{det}(A) = a(-4) - 2(-36) + 3(29) \]

### Step 3: Simplify the determinant
Now, simplifying the expression:

\[ \text{det}(A) = -4a + 72 + 87 \]
\[ \text{det}(A) = -4a + 159 \]

### Step 4: Set the determinant to be non-zero
For the system to have a unique solution, the determinant must be non-zero:

\[ -4a + 159 \neq 0 \]

### Step 5: Solve for \( a \)
Now, we solve for \( a \):

\[ -4a \neq -159 \]
\[ a \neq \frac{159}{4} \]

### Conclusion
Thus, the value of \( a \) such that the equations have a unique solution is:

\[ a \in \mathbb{R} \setminus \left\{ \frac{159}{4} \right\} \]