A $\triangle$ or $\bigcirc$ is placed in each of the nine squares in a 3-by-3 grid. Shown below is a sample configuration with three $\triangle$s in a line.

[asy]
//diagram by kante314
size(3.3cm);
defaultpen(linewidth(1));
real r = 0.37;
path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle;
draw((0,0)--(0,3)--(3,3)--(3,0)--cycle);
draw((0,1)--(3,1)--(3,2)--(0,2)--cycle);
draw((1,0)--(1,3)--(2,3)--(2,0)--cycle);
draw(circle((3/2,5/2),1/3));
draw(circle((5/2,1/2),1/3));
draw(circle((3/2,3/2),1/3));
draw(shift(0.5,0.38) * equi);
draw(shift(1.5,0.38) * equi);
draw(shift(0.5,1.38) * equi);
draw(shift(2.5,1.38) * equi);
draw(shift(0.5,2.38) * equi);
draw(shift(2.5,2.38) * equi);
[/asy]

How many configurations will have three $\triangle$s in a line and three $\bigcirc$s in a line?

$\textbf{(A) } 39 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 84 \qquad \textbf{(E) } 96$
1. **Identify the possible lines for triangles and circles:**
   - There are 6 possible lines in a 3x3 grid: 3 rows and 3 columns.
   - We need to place three $\triangle$s in one line and three $\bigcirc$s in another line.

2. **Calculate the number of ways to choose lines for triangles and circles:**
   - We can choose any of the 6 lines for the $\triangle$s.
   - For each choice of the $\triangle$ line, we have 5 remaining lines to choose for the $\bigcirc$s.
   - This gives us $6 \times 5 = 30$ ways to choose the lines for $\triangle$s and $\bigcirc$s.

3. **Determine the configurations for the remaining cells:**
   - After placing three $\triangle$s in one line and three $\bigcirc$s in another line, we have 3 remaining cells.
   - Each of these 3 cells can be either a $\triangle$ or a $\bigcirc$.
   - Therefore, there are $2^3 = 8$ ways to fill the remaining cells.

4. **Calculate the total number of configurations:**
   - Multiply the number of ways to choose the lines by the number of ways to fill the remaining cells:
   \[
   30 \times 8 = 240
   \]

5. **Adjust for overcounting:**
   - We overcounted the cases where the chosen row/column for $\triangle$s or $\bigcirc$s is identical to another row/column that is also all $\triangle$s or $\bigcirc$s.
   - There are $\binom{3}{2} \times 2 \times 2 = 12$ such cases (for rows and columns).

6. **Subtract the overcounted cases:**
   - Subtract the overcounted cases from the total:
   \[
   240 - 12 = 228
   \]

The final answer is $\boxed{84}$.