Illustration of the vector equation of a plane:
Like we can see from the figure upside, a plane can be expressed in the following form as an identity:
$\vec A - \vec Q = s(\vec P -\vec Q) + t(\vec R - \vec Q)$
And, so:
$\vec A = s(\vec P -\vec Q) + t(\vec R - \vec Q) + \vec Q $
This is the vector equation of a plane. $s$ And $t$ in here are unknowns, and the author of the book i have read notes that if we count the coordinates of vectors as knowns, we can eliminate the unknowns and obtain the factors of the plane equation ($a, b, c$ and $d$ in the $ax + by + cz + d = 0$) as:
$a = (y_R - y_Q)(z_P - z_Q) - (z_R - z_Q)(y_P - y_Q)$
$b = (z_R - z_Q)(x_P - x_Q) - (x_R - x_Q)(z_P - z_Q)$
$c = (x_R - x_Q)(y_P - y_Q) - (y_R - y_Q)(x_P - x_Q)$
$d = -(ax_Q + by_Q + cz_Q)$
Three noncolinear points define a plane, so one way to perform this conversion is to plug in $\vec Q$, $\vec Q+\vec P$ and $\vec Q+\vec R$ into the generic plane equation and solve for the coefficients. However, there’s a more direct and much less tedious way to obtain the result.
I hesitate to call the parametric equation that you’re starting from the vector equation of a plane because there’s an implicit vector equation that is quite relevant to your question. This other vector equation is known variously as the point-normal form or Hesse normal form. It is $$\vec N\cdot(\vec X -\vec P) = 0$$ or $$\vec N\cdot\vec X+d=0.$$ (You’ll sometimes see the latter form written with $-d$ instead.) Here $\vec N$ is a vector normal (perpendicular) to the plane and $\vec P$ is any fixed known point on it. If you expand the second form of this equation in terms of coordinates, you get exactly the implicit Cartesian equation of the plane that you’re looking for, with $\vec N = (a,b,c)$ and $d = -\vec N\cdot\vec P = -(a,b,c)\cdot\vec P$. Geometrically, this equation says that the plane is the set of points whose orthogonal projections onto $\vec N$ are equal to some fixed value. You can see how this works if you add to the illustration in the question a vector $\vec N$ that goes from $O$ to the plane and is perpendicular to it.
You have a known vector on the plane, $\vec Q$, so all you need to do is to find a suitable $\vec N$. For this, use the property that the cross product of two vectors is orthogonal to them both, so a convenient choice is $$\vec N = (\vec R-\vec Q)\times(\vec P-\vec Q).$$ If you expand this in coordinates, you’ll get the expressions in your question.