Let $k$ be an algebraically closed field, and $a$ an ideal of the polynomial ring $k[x_1,x_2,\dots,x_n]$. The strong form of Hilbert's Nullstellensatz says that $I(Z(a))=\sqrt{a}$.
Note:- Initially, the question was regarding $I(\langle (x-1)+(y-2)+(z-3)\rangle$. I then replaced $(x-1)+(y-2)+(z-3)$ with $(x-1)^2+(y-2)^2+(z-3)^2$ so as to ensure that the zero locus consists of only the point $(1,2,3)$.
I have a question regarding this statement. Take the polynomial ring $\Bbb{C}(x,y,z)$, where $\Bbb{C}$ is obviously an algebraically closed field. Now take the point $(1,2,3)\in\Bbb{C^3}$. Clearly $\langle (x-1)^2+(y-2)^2+(z-3)^2\rangle$ has $(1,2,3)$ as its only zero. In other words, $Z(\langle (x-1)^2+(y-2)^2+(z-3)^2\rangle)=(1,2,3)$. Now let us consider $I(Z(\langle (x-1)^2+(y-2)^2+(z-3)^2\rangle))$. This will include the polynomial $(x-1)+(y-2)$, among others like $(x-1)$ and $(y-2)$. However, $(x-1)+(y-2)\notin \sqrt{\langle (x-1)^2+(y-2)^2+(z-3)^2\rangle}$. Isn't this a contradiction?
Note that the equation $x + y + z = 6$ defines a plane in $\mathbb{C}^3$; in particular, $(1, 2, 3)$ is not the only point which satisfies this equation. For example, $(6, 0, 0)$ lies in the plane, so $$(6, 0, 0) \in Z(\langle x + y + z - 6\rangle).$$ Note that the polynomial $(x - 1) + (y - 2)$ is not zero at $(6, 0, 0)$, so $$(x - 1) + (y -2) \not\in I(Z(\langle x + y + z - 6\rangle)).$$ In particular, there is no contradiction.