Let $f(t) = \begin{vmatrix} \cos t & 1 & 1 \\ 2\sin t & 2t & 1 \\ \sin t & t & t \end{vmatrix}$, then $\displaystyle \lim_{t \to 0}\dfrac{f(t)}{t^2}$ is equal to:

Expand the determinant using first row: $f(t) = \cos t \begin{vmatrix} 2t & 1 \\ t & t \end{vmatrix} - 1 \begin{vmatrix} 2\sin t & 1 \\ \sin t & t \end{vmatrix} + 1 \begin{vmatrix} 2\sin t & 2t \\ \sin t & t \end{vmatrix}$ Simplify and expand around $t \to 0$ using $\sin t \approx t$ and $\cos t \approx 1 - t^2/2$. After simplification, $\dfrac{f(t)}{t^2} \to 3$. $\boxed{\text{Answer: (D) 3}}$