For a certain value of $k$, the system
\begin{align*}
x + y + 3z &= 10, \\
x - 2y + 8z &= 7, \\
kx + 5z &= 3<#> \end{align*}<#> has no solutions. What is this value of $k$?
Subtracting the second equation from the first, we get
\begin{align*}
3y - 5z &= 3.
\end{align*}Adding $3y - 5z = 3$ to the third equation, we get
\[(3y - 5z) + kx + 5z = 6,\]which simplifies to $3y + kx = 6.$ Rearranging, we get $kx = -3y + 6.$ Then $x = -\frac{3}{k} y + \frac{6}{k}$. Substituting into $x + y + 3z = 10,$ we get
\[-\frac{3}{k} y + y + 3z = 10.\]Hence,
\[3z = -\left( \frac{3}{k} - 1 \right) y + 10,\]so $y = \frac{k}{k - 3} z + \frac{10}{k - 3}.$ Substituting into $x = -\frac{3}{k} y + \frac{6}{k},$ we get
\[x = -\frac{3k}{k - 3} z - \frac{6k}{k - 3}.\]Then
\[\begin{cases} x = -\frac{3k}{k - 3} z - \frac{6k}{k - 3}, \\ y = \frac{k}{k - 3} z + \frac{10}{k - 3}.\end{cases}\]We can write this as
\[\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -\frac{3k}{k - 3} \\ \frac{k}{k - 3} \\ 1 \end{pmatrix} z + \begin{pmatrix} -\frac{6k}{k - 3} \\ \frac{10}{k - 3} \\ 0 \end{pmatrix}.\]For this system to have no solutions, the vectors $\begin{pmatrix} -\frac{3k}{k - 3} \\ \frac{k}{k - 3} \\ 1 \end{pmatrix}$ and $\begin{pmatrix} -\frac{6k}{k - 3} \\ \frac{10}{k - 3} \\ 0 \end{pmatrix}$ must be parallel (because then the geometric interpretation is that two planes corresponding to those vectors are coincident, but the system does not have a solution). Note that the proportionality constant cannot be 0 (because $\begin{pmatrix} -\frac{3k}{k - 3} \\ \frac{k}{k - 3} \\ 1 \end{pmatrix}$ and $\begin{pmatrix} -\frac{6k}{k - 3} \\ \frac{10}{k - 3} \\ 0 \end{pmatrix}$ are nonzero vectors). Thus,
\[\frac{-\frac{3k}{k - 3}}{-\frac{6k}{k - 3}} = \frac{\frac{k}{k - 3}}{\frac{10}{k - 3}}.\]This reduces to $-k = \frac{1}{5}.$ Hence, $k = \boxed{-\frac{1}{5}}.$