Let $L$ be the line in space that passes through the origin and the point $(2,1,-2).$  Find the reflection of the point $(3,6,15)$ across $L.$
Let $\mathbf{v} = \begin{pmatrix} 3 \\ 6 \\ 15 \end{pmatrix}$ and $\mathbf{w} = \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}.$

[asy]
import three;

size(180);
currentprojection = perspective(6,3,2);

triple I = (1,0,0), J = (0,1,0), K = (0,0,1), O = (0,0,0);
triple V = (3,2,2), W = (4,1,3), P = dot(V,W)/abs(W)^2*W, R = 2*P - V;

draw(V--R,dashed);
draw(0.85*P--(0.85*P + 0.15*(V - P))--(P + 0.15*(V - P)));
draw(O--V,Arrow3(6));
draw(P--W,Arrow3(6));
draw(O--P,Arrow3(6));
draw(O--R,Arrow3(6));
draw(O--3*I, Arrow3(6));
draw(O--3*J, Arrow3(6));
draw(O--3*K, Arrow3(6));

label("$x$", 3.2*I);
label("$y$", 3.2*J);
label("$z$", 3.2*K);
label("$\mathbf{v}$", V, NE);
label("$\mathbf{w}$", W, N);
label("$\mathbf{p}$", P, SW);
label("$\mathbf{r}$", R, SW);
[/asy]

Let $\mathbf{p}$ be the projection of $\mathbf{v}$ onto $\mathbf{w},$ so
\[\mathbf{p} = \frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}} \mathbf{w} = \frac{\begin{pmatrix} 3 \\ 6 \\ 15 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}}{\begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}} \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix} = \frac{-18}{9} \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} -4 \\ -2 \\ 4 \end{pmatrix}.\]Let $\mathbf{r}$ be the reflection of $\mathbf{v}$ across line $L.$  Then $\mathbf{p}$ is the midpoint of $\mathbf{v}$ and $\mathbf{r},$ so
\[\mathbf{p} = \frac{\mathbf{v} + \mathbf{r}}{2}.\]Then
\[\mathbf{r} = 2 \mathbf{p} - \mathbf{v} = 2 \begin{pmatrix} -4 \\ -2 \\ 4 \end{pmatrix} - \begin{pmatrix} 3 \\ 6 \\ 15 \end{pmatrix} = \begin{pmatrix} -11 \\ -10 \\ -7 \end{pmatrix}.\]Hence, the resulting point is $\boxed{(-11,-10,-7)}.$