This is what I have done:
Let the centre of the sphere be origin for simplicity.
Let r be the radius of the sphere.
So, it's equation is $x^2 + y^2 +z^2 = r^2$
Let the point $(x_1,y_1,z_1)$ be the point of tangency on the sphere.
Let the given point be $(p_1,p_2,p_3)$.
Then the equation of the tangent planes is given by:
$x\times x_1 + y\times y_1 + z\times z_1 = r^2$
The point $(p_1,p_2,p_3)$ satisfies this equation and $(x_1,y_1,z_1)$ satisfies the sphere's equation which gives two equations. Please help on how to proceed further. I think that there will be only one parameter in the final equation representing all the planes.
You wrote that “$(p_1,p_2,p_3)$ satisfies this equation”. Why? You must impose that, that is, $(x_1,y_1,z_1)$ must be chosen in such a way that $p_1x_1+p_2y_1+p_3z_1=r^2$. The set of all points that satisfy this equality is a plane and when you intersect this plane with the sphere you get a circle. The points of that circle are precisely the points of the sphere such that the tangent plane to the sphere passing through that point contains the point $(p_1,p_2,p_3)$.
So, the answer is: the planes tangent to the sphere which contain $(p_1,p_2,p_3)$ are the planes of the form $x\times x_1+y\times y_1+z\times z_1=r^2$, where $(x_1,y_1,z_1)$ is such that $p_1x_1+p_2y_1+p_3z_1=r^2$.