Assume $p1=(x_1,y_1,z_1)$ , $p2=(x_2,y_2,z_2)$ and $p3=(x_3,y_3,z_3)$
THIS IS HOW I IMAGINE IT (so it may be wrong)
$p_1p_2^\rightarrow =(x_2-x_1,y_2-y_1,z_2-z_1)$ which is a vector on the plane
i tried to get the projection vector of $p_1p_3^\rightarrow$ onto $p_1p_2^\rightarrow$ then the rejection Which as i think so is the normal vector $n^\rightarrow$ according to https://en.wikipedia.org/wiki/Vector_projection
states that the vector rejection is the orthogonal projection of $p_1p_3^\rightarrow$ onto the plane
vector projection of $p_1p_3^\rightarrow$ onto $p_1p_2^\rightarrow =$ $\frac{p_1p_3^\rightarrow.p_1p_2^\rightarrow}{\lVert p_1p_2^\rightarrow\rVert^2}p_1p_2^\rightarrow=v^\rightarrow$
then the Vector rejection which is the normal vector $n^\rightarrow$ =$p_1p_3^\rightarrow - v^\rightarrow$
then the equation of the plane is $$n^\rightarrow.((x,y,z)-(x_1,y_1,z_1))=0$$
i don't know if this is right or wrong because as you see i didn't use the distance he gave and i searched a lot but i found nothing
so i need any clarification please
Calling $p^*$ the tangent point between the plane and the sphere $||p-p_3||= d$, we have the relations
$$ \cases{ ||p^*-p_3||^2=d^2\\ (p^*-p_3)\cdot(p_1-p_2) = 0\\ (p^*-p_3)\cdot(p_1-p^*) = 0 } $$
We have three conditions and three unknowns which are the $p^*$ components.
Case study
$$ \cases{ p_1 = (1,0,1)\\ p_2 = (2,-1,0)\\ p_3 = (-2,1,-2)\\ d = 2 } $$
$$ p^* = \left(-\frac{2}{7} \left(5+\sqrt{11}\right),-\frac{3}{7} \left(\sqrt{11}-2\right),\frac{1}{7} \left(\sqrt{11}-9\right)\right) $$
and the plane equation is
$$ (p-p^*)\cdot(p^*-p_3)=0,\ \ \ p=(x,y,z) $$