Let \P(x_{1},y_{1})\ and \Q(x_{2},y_{2})\,

By solving \ \begin{cases} (x-3)^{2}+(y+4)^{2}=4 \\newline y=kx \end{cases}\ and eliminating \y\, we get \(k^{2}+1)x^{2}+(8k-6)x+21=0\,

Thus, \x_{1}+x_{2}= \dfrac {-8k+6}{k^{2}+1}\ and \x_{1}x_{2}= \dfrac {21}{k^{2}+1}\.

From this, we can find \y_{1}y_{2}=kx_{1}⋅kx_{2}=k^{2}x_{1}x_{2}\,

Therefore, \\overrightarrow{OP}⋅ \overrightarrow{OQ}=x_{1}x_{2}+y_{1}y_{2}=(1+k^{2})x_{1}x_{2}=(1+k^{2})⋅ \dfrac {21}{k^{2}+1}=21\.

Since \\overrightarrow{OP}\ and \\overrightarrow{OQ}\ are collinear and in the same direction, \|OP|⋅|OQ|= \overrightarrow{OP}⋅ \overrightarrow{OQ}=21\.

Therefore, the correct choice is: \\boxed{D}\

By setting \P(x_{1},y_{1})\ and \Q(x_{2},y_{2})\, solving the equation of the line and the circle together to eliminate \y\, and organizing it into a quadratic equation in terms of \x\, we use the relationship between the roots and coefficients to calculate the expression of \x_{1}x_{2}\ in terms of \k\, and then find the expression of \y_{1}y_{2}\ in terms of \k\. Using the formula for the dot product of vectors, we can find the answer.

This problem examines the relationship between a line and a circle, the operation and properties of the dot product of planar vectors, and is considered a medium-difficulty question. It also tests the mathematical transformation thought and overall operation thought. In problems involving the intersection points of a line and a circle, the method of setting without solving is often used to simplify the problem.