(1 point) Find the intersection of the lines $r(t)=\langle 6-2 t,-1+2 t,-10+8 t\rangle$ and $\mathbf{R}(s)=\langle 13+5 s,-8-5 s, 4+s\rangle$. Write your answer as point $(a, b, c)$ where $a, b$, and $c$ are numbers. Answer: Note: If there is no intersection, write "none".

Step 1 ：The intersection of two lines occurs when their parametric equations are equal. That is, we need to find a value of t and s such that: \(6 - 2t = 13 + 5s\), \(-1 + 2t = -8 - 5s\), \(-10 + 8t = 4 + s\).

Step 2 ：From the first equation, we can express s in terms of t: \(6 - 2t = 13 + 5s\), \(5s = -7 + 2t\), \(s = (-7 + 2t) / 5\).

Step 3 ：Substitute s into the second equation: \(-1 + 2t = -8 - 5((-7 + 2t) / 5)\), \(-1 + 2t = -8 + 7 - 2t\), \(4t = 1\), \(t = 1 / 4\).

Step 4 ：Substitute t = 1 / 4 into the third equation: \(-10 + 8(1 / 4) = 4 + s\), \(-10 + 2 = 4 + s\), \(s = -12 + 2\), \(s = -10\).

Step 5 ：Substitute t = 1 / 4 and s = -10 into the parametric equations of the lines: \(r(t) = <6 - 2(1 / 4), -1 + 2(1 / 4), -10 + 8(1 / 4)> = <5, -0.5, -8>\), \(R(s) = <13 + 5(-10), -8 - 5(-10), 4 - 10> = <-37, 42, -6>\).

Step 6 ：Since \(r(t) \neq R(s)\), the lines do not intersect. Therefore, the answer is \(\boxed{\text{none}}\).