Which of the following has`//`have value equal to zero ?

Options

A. `|{:(8,,2,,7),(12,,3,,5),(16,,4,,3):}|`
B. `|{:(1//a,,a^(2),,bc),(1//b,,b^(2),,ac),(1//c,,c^(2),,ab):}|`
C. `|{:(a+b,,2a+b,,3a+b),(2a+b,,3a+b,,4a+b),(4a+b,,5a+b,,6a+b):}|`
D. `|{:(2,,43,,6),(7,,35,,4),(3,,17,,2):}|`

Explanation

To determine which of the given determinants has a value equal to zero, we will analyze each option step by step.

### Step 1: Analyze the First Determinant
The first determinant is:
\[ \begin{vmatrix} 8 & 12 & 16 \\ 2 & 3 & 4 \\ 7 & 5 & ? \end{vmatrix} \]

**Transformation:**
We will perform row operations:
- Change \( R_3 \) to \( R_3 - R_2 \)
- Change \( R_2 \) to \( R_2 - R_1 \)

After performing these operations, we have:
\[ R_1 = (8, 12, 16) \\ R_2 = (2 - 8, 3 - 12, 4 - 16) = (-6, -9, -12) \\ R_3 = (7 - 2, 5 - 3, ?) = (5, 2, ?) \]

Now, we notice that after simplification, two rows become identical, leading to:
\[ \begin{vmatrix} 8 & 12 & 16 \\ 0 & 0 & 0 \\ 5 & 2 & ? \end{vmatrix} \]

Since two rows are identical, the value of this determinant is **0**.

### Step 2: Analyze the Second Determinant
The second determinant is:
\[ \begin{vmatrix} \frac{1}{a} & \frac{1}{b} & \frac{1}{c} \\ \frac{1}{b} & \frac{1}{c} & a^2 \\ b^2 & c^2 & a^2 \\ \end{vmatrix} \]

**Transformation:**
Multiply the first row by \( a \), the second row by \( b \), and the third row by \( c \):
\[ \begin{vmatrix} 1 & \frac{a}{b} & \frac{a}{c} \\ \frac{b}{b} & \frac{b}{c} & a^2 \\ bc & c^2 & a^2 \\ \end{vmatrix} \]

After simplification, we notice that two columns become identical, leading to:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & b^3 & 1 \\ bc & c^2 & a^2 \\ \end{vmatrix} \]

Since two columns are identical, the value of this determinant is **0**.

### Step 3: Analyze the Third Determinant
The third determinant is:
\[ \begin{vmatrix} a + b & 2a + b & 4a + b \\ 2a + b & 3a + b & 5a + b \\ 4a + b & 6a + b & ? \\ \end{vmatrix} \]

**Transformation:**
Change \( R_2 \) to \( R_2 - R_1 \) and \( R_3 \) to \( R_3 - R_2 \):
\[ \begin{vmatrix} a + b & 2a + b & 4a + b \\ 0 & a & a \\ 0 & 0 & ? \\ \end{vmatrix} \]

After simplification, we notice that two rows become identical, leading to:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & ? \\ \end{vmatrix} \]

Since two rows are identical, the value of this determinant is **0**.

### Step 4: Analyze the Fourth Determinant
The fourth determinant is:
\[ \begin{vmatrix} 17 & 6 & 4 \\ 2 & 6 & 4 \\ 2 & 6 & 4 \\ \end{vmatrix} \]

**Transformation:**
Change \( C_2 \) to \( C_2 - 7C_3 \):
\[ \begin{vmatrix} 17 & 6 - 28 & 4 \\ 2 & 6 - 28 & 4 \\ 2 & 6 - 28 & 4 \\ \end{vmatrix} \]

After simplification, we find that the determinant does not yield identical rows or columns, and we calculate:
\[ = 17(6 - 4) - 6(2 - 4) + 4(2 - 2) = 2 \]
Thus, the value of this determinant is **not 0**.

### Final Conclusion
The determinants that have a value equal to zero are the first three options.