Find $x+y+z$ when $$a_1x+a_2y+a_3z= a$$$$b_1x+b_2y+b_3z=b$$$$c_1x+c_2y+c_3z=c$$ Given that $$a_1\left(b_2c_3-b_3c_2\right)-a_2\left(b_1c_3-b_3c_1\right)+a_3\left(b_1c_2-b_2c_1\right)=9$$$$a\left(b_2c_3-b_3c_2\right)-a_2\left(bc_3-b_3c\right)+a_3\left(bc_2-b_2c\right)=17$$$$a_1\left(bc_3-b_3c\right)-a\left(b_1c_3-b_3c_1\right)+a_3\left(b_1c-bc_1\right)=-8$$$$a_1\left(b_2c-bc_2\right)-a_2\left(b_1c-bc_1\right)+a\left(b_1c_2-b_2c_1\right)=7.$$

[i]2017 CCA Math Bonanza Lightning Round #5.1[/i]
1. We start with the given system of linear equations:
   \[
   \begin{cases}
   a_1x + a_2y + a_3z = a \\
   b_1x + b_2y + b_3z = b \\
   c_1x + c_2y + c_3z = c
   \end{cases}
   \]
   We are also given the determinants:
   \[
   \left| \begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array} \right| = 9
   \]
   \[
   \left| \begin{array}{ccc} a & b & c \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array} \right| = 17
   \]
   \[
   \left| \begin{array}{ccc} a_1 & b_1 & c_1 \\ a & b & c \\ a_3 & b_3 & c_3 \end{array} \right| = -8
   \]
   \[
   \left| \begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a & b & c \end{array} \right| = 7
   \]

2. Using Cramer's Rule, we can solve for \(x\), \(y\), and \(z\). According to Cramer's Rule:
   \[
   x = \frac{\left| \begin{array}{ccc} a & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array} \right|}{\left| \begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array} \right|}
   \]
   \[
   y = \frac{\left| \begin{array}{ccc} a_1 & b_1 & c_1 \\ a & b & c \\ a_3 & b_3 & c_3 \end{array} \right|}{\left| \begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array} \right|}
   \]
   \[
   z = \frac{\left| \begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a & b & c \end{array} \right|}{\left| \begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array} \right|}
   \]

3. Substituting the given determinants:
   \[
   x = \frac{17}{9}
   \]
   \[
   y = \frac{-8}{9}
   \]
   \[
   z = \frac{7}{9}
   \]

4. Adding \(x\), \(y\), and \(z\):
   \[
   x + y + z = \frac{17}{9} + \frac{-8}{9} + \frac{7}{9} = \frac{17 - 8 + 7}{9} = \frac{16}{9}
   \]

The final answer is \(\boxed{\frac{16}{9}}\)