Derivative of cross product equation

You are quite right. You can't sensibly equate a vector to a scalar. So the scalar function of $t$ on the right hand side of your equation can't be right. But you could have a vector function of $t$.

The cross product of two planar vectors is a scalar.

$$ pmatrix{a b} times pmatrix{x y} = a y - b x $$

Also, note the following 2 planar cross products that exist between a vector and a scalar (out of plane vector).

$$ pmatrix{a b} times omega = pmatrix{omega, b -omega, a} $$

$$ omega times pmatrix{x y} = pmatrix{-omega, y omega , x} $$

All of the above are planar projections of the one 3D cross product.

Also note that the derivative is a linear operator which means the product rule applies

$$ tfrac{rm d}{{rm d}t} ( vec{a} times vec{b} ) = ( tfrac{rm d}{{rm d}t} vec{a} ) times vec{b} + vec{a} times (tfrac{rm d}{{rm d}t} vec{b}) $$

The vectors P and Q define a plane. The cross product is perpendicular to that plane. Your time function could describe the motion of an object in that perpendicular direction.