If `a!=6,b,c` satisfy`|[a,2b,2c],[3,b,c],[4,a,b]|=0` ,then abc =

Options

A. `a +b +c`
B. 0
C. `b^(3)`
D. `ab + bc`

Answer

To solve the determinant problem given by the equation

\[ \begin{vmatrix} a & 2b & 2c \\ 3 & b & c \\ 4 & a & b \end{vmatrix} = 0 \]

we will follow these steps:

### Step 1: Write the determinant
The determinant can be expressed as:

\[ D = a \begin{vmatrix} b & c \\ a & b \end{vmatrix} - 2b \begin{vmatrix} 3 & c \\ 4 & b \end{vmatrix} + 2c \begin{vmatrix} 3 & b \\ 4 & a \end{vmatrix} \]

### Step 2: Calculate the 2x2 determinants
Calculating the 2x2 determinants:

1. \(\begin{vmatrix} b & c \\ a & b \end{vmatrix} = b^2 - ac\)

2. \(\begin{vmatrix} 3 & c \\ 4 & b \end{vmatrix} = 3b - 4c\)

3. \(\begin{vmatrix} 3 & b \\ 4 & a \end{vmatrix} = 3a - 4b\)

### Step 3: Substitute back into the determinant
Now substituting these values back into the expression for \(D\):

\[ D = a(b^2 - ac) - 2b(3b - 4c) + 2c(3a - 4b) \]

### Step 4: Expand the determinant
Expanding this, we have:

\[ D = ab^2 - a^2c - 6b^2 + 8bc + 6ac - 8bc \]

This simplifies to:

\[ D = ab^2 - a^2c - 6b^2 + 6ac = 0 \]

### Step 5: Factor the equation
We can factor this equation:

\[ D = (a - 6)(b^2 - ac) = 0 \]

### Step 6: Analyze the factors
From the factorization, we have two cases:

1. \(a - 6 = 0\) which gives \(a = 6\), but we know \(a \neq 6\).
2. \(b^2 - ac = 0\) which gives \(b^2 = ac\).

### Step 7: Find the value of \(abc\)
From \(b^2 = ac\), we can express \(abc\):

\[ abc = b \cdot b^2 = b \cdot ac = b^3 \]

### Conclusion
Thus, the value of \(abc\) is \(b^3\).