The sum of roots of the equations
`|{:(x+2,,2x+3,,3x+4),(2x+3,,3x+4,,4x+5),(3x+5,,5x+8,,10x+17):}| =0 " is " "____"`

Explanation

To solve the problem, we need to find the sum of the roots of the determinant equation given by:

\[ \left| \begin{array}{ccc} x + 2 & 2x + 3 & 3x + 4 \\ 2x + 3 & 3x + 4 & 4x + 5 \\ 3x + 5 & 5x + 8 & 10x + 17 \end{array} \right| = 0 \]

### Step-by-Step Solution:

**Step 1: Apply Row Operations**
We can simplify the determinant by performing row operations. We will replace \( R_2 \) with \( R_2 - R_1 \) and \( R_3 \) with \( R_3 - R_2 \).

\[ \Rightarrow \left| \begin{array}{ccc} x + 2 & 2x + 3 & 3x + 4 \\ (2x + 3) - (x + 2) & (3x + 4) - (2x + 3) & (4x + 5) - (3x + 4) \\ (3x + 5) - (2x + 3) & (5x + 8) - (3x + 4) & (10x + 17) - (4x + 5) \end{array} \right| \]

Calculating the new rows gives us:

\[ \Rightarrow \left| \begin{array}{ccc} x + 2 & 2x + 3 & 3x + 4 \\ x + 1 & x + 1 & x + 1 \\ x + 2 & 2x + 4 & 6x + 12 \end{array} \right| \]

**Step 2: Factor Out Common Terms**
We can factor out \( (x + 1) \) from the second row and \( (x + 2) \) from the third row:

\[ = (x + 1)(x + 2) \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 6 \end{array} \right| \]

**Step 3: Calculate the 2x2 Determinant**
Now we need to calculate the determinant of the 2x2 matrix:

\[ \left| \begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array} \right| = (1)(2) - (1)(1) = 2 - 1 = 1 \]

**Step 4: Substitute Back**
Thus, we have:

\[ (x + 1)(x + 2) \cdot 1 = 0 \]

**Step 5: Find the Roots**
Setting the equation to zero gives us:

\[ (x + 1)(x + 2) = 0 \]

The roots are:

\[ x + 1 = 0 \Rightarrow x = -1 \\ x + 2 = 0 \Rightarrow x = -2 \]

**Step 6: Sum of the Roots**
The sum of the roots is:

\[ -1 + (-2) = -3 \]

### Final Answer:
The sum of the roots of the equation is:

\[ \text{Sum of roots} = -3 \]