Answer
See proof
Work Step by Step
\begin{align*}
\ell_1: \quad & x_1 = 2+8t_1, \quad y_1=6-8t_1, \quad z_1=10t_1 \\
\ell_2: \quad & x_2 = 3+8t_2, \quad y_2=5-3t_2, \quad z_2=6+t_2 \
\end{align*}
To check if the lines are skew, we can see if there exists a common value for $t_1$ and $t_2$ such that the lines intersect. We can equate the respective parametric equations of the two lines and solve for $t_1$ and $t_2$:
\begin{align*}
x_1 = x_2 \quad & \Rightarrow \quad 2+8t_1 = 3+8t_2 \\
y_1 = y_2 \quad & \Rightarrow \quad 6-8t_1 = 5-3t_2 \\
z_1 = z_2 \quad & \Rightarrow \quad 10t_1 = 6+t_2
\end{align*}
Solving this system of equations yields $t_1 = 1$, $t_2 = -1$, and $z_1 = 10$ (which satisfies the third equation). Since the lines do not intersect, we can conclude that they are skew.
The direction vectors of the two lines are $\mathbf{v_1} = \langle 8, -8, 10 \rangle$ and $\mathbf{v_2} = \langle 8, -3, 1 \rangle$, which are not parallel. Therefore, the two lines are non-coplanar and skew.
