Solve given system of equations by matrix method:
`(2)/(a)+(3)/(b)+(4)/(c)=-3,(5)/(a)+(4)/(b)-(6)/(c)=4,(3)/(a)-(2)/(b)-(2)/(c)=6`

Answer

To solve the given system of equations using the matrix method, we start by rewriting the equations in a suitable form. The equations are:

1. \(\frac{2}{a} + \frac{3}{b} + \frac{4}{c} = -3\)
2. \(\frac{5}{a} + \frac{4}{b} - \frac{6}{c} = 4\)
3. \(\frac{3}{a} - \frac{2}{b} - \frac{2}{c} = 6\)

Let us define:
- \(x = \frac{1}{a}\)
- \(y = \frac{1}{b}\)
- \(z = \frac{1}{c}\)

Now we can rewrite the equations in terms of \(x\), \(y\), and \(z\):

1. \(2x + 3y + 4z = -3\)
2. \(5x + 4y - 6z = 4\)
3. \(3x - 2y - 2z = 6\)

### Step 1: Form the Matrix Equation
We can express the above equations in matrix form \(AX = B\), where:

\[ A = \begin{bmatrix} 2 & 3 & 4 \\ 5 & 4 & -6 \\ 3 & -2 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} -3 \\ 4 \\ 6 \end{bmatrix} \]

### Step 2: Calculate the Determinant of Matrix \(A\)
To find \(A^{-1}\), we first need to calculate the determinant of \(A\):

\[ \text{det}(A) = 2 \begin{vmatrix} 4 & -6 \\ -2 & -2 \end{vmatrix} - 3 \begin{vmatrix} 5 & -6 \\ 3 & -2 \end{vmatrix} + 4 \begin{vmatrix} 5 & 4 \\ 3 & -2 \end{vmatrix} \]

Calculating the minors:

1. \(\begin{vmatrix} 4 & -6 \\ -2 & -2 \end{vmatrix} = (4)(-2) - (-6)(-2) = -8 - 12 = -20\)

2. \(\begin{vmatrix} 5 & -6 \\ 3 & -2 \end{vmatrix} = (5)(-2) - (-6)(3) = -10 + 18 = 8\)

3. \(\begin{vmatrix} 5 & 4 \\ 3 & -2 \end{vmatrix} = (5)(-2) - (4)(3) = -10 - 12 = -22\)

Now substituting back into the determinant formula:

\[ \text{det}(A) = 2(-20) - 3(8) + 4(-22) = -40 - 24 - 88 = -152 \]

### Step 3: Calculate the Adjoint of Matrix \(A\)
Next, we need to find the cofactors of each element in \(A\) to form the adjoint matrix.

Calculating the cofactors:

\[ C_{11} = \begin{vmatrix} 4 & -6 \\ -2 & -2 \end{vmatrix} = -20 \quad \Rightarrow \quad C_{11} = -20 \]

\[ C_{12} = -\begin{vmatrix} 5 & -6 \\ 3 & -2 \end{vmatrix} = -8 \quad \Rightarrow \quad C_{12} = 8 \]

\[ C_{13} = \begin{vmatrix} 5 & 4 \\ 3 & -2 \end{vmatrix} = -22 \quad \Rightarrow \quad C_{13} = -22 \]

Continuing this process for all elements, we find:

\[ \text{adj}(A) = \begin{bmatrix} -20 & 8 & -22 \\ -2 & -16 & 13 \\ -34 & 32 & -7 \end{bmatrix} \]

### Step 4: Calculate the Inverse of Matrix \(A\)
Now we can find \(A^{-1}\):

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{-152} \begin{bmatrix} -20 & 8 & -22 \\ -2 & -16 & 13 \\ -34 & 32 & -7 \end{bmatrix} \]

### Step 5: Solve for \(X\)
Now we can calculate \(X\):

\[ X = A^{-1}B = \frac{1}{-152} \begin{bmatrix} -20 & 8 & -22 \\ -2 & -16 & 13 \\ -34 & 32 & -7 \end{bmatrix} \begin{bmatrix} -3 \\ 4 \\ 6 \end{bmatrix} \]

Calculating the product:

1. First row: \((-20)(-3) + (8)(4) + (-22)(6) = 60 + 32 - 132 = -40\)
2. Second row: \((-2)(-3) + (-16)(4) + (13)(6) = 6 - 64 + 78 = 20\)
3. Third row: \((-34)(-3) + (32)(4) + (-7)(6) = 102 + 128 - 42 = 188\)

Thus,

\[ X = \frac{1}{-152} \begin{bmatrix} -40 \\ 20 \\ 188 \end{bmatrix} = \begin{bmatrix} \frac{40}{152} \\ -\frac{20}{152} \\ -\frac{188}{152} \end{bmatrix} \]

### Step 6: Find \(a\), \(b\), and \(c\)
Now we convert back to \(a\), \(b\), and \(c\):

1. \(x = \frac{1}{a} = \frac{40}{152} \Rightarrow a = \frac{152}{40} = 3.8\)
2. \(y = \frac{1}{b} = -\frac{20}{152} \Rightarrow b = -\frac{152}{20} = -7.6\)
3. \(z = \frac{1}{c} = -\frac{188}{152} \Rightarrow c = -\frac{152}{188} = -0.807\)

### Final Answer
Thus, the solution to the system of equations is:

\[ a = 3.8, \quad b = -7.6, \quad c = -0.807 \]