I am getting the contradiction described below. I have been racking my brain since yesterday to figure out what the problem is here, but I just can't. I must be making some super silly mistake. Please help me figure it out.
$\textbf{Notation:}$ For any ideal $A \subseteq k[x_1,...,x_n]$, denote the variety of $A$ by $V(A)$. For any variety $V$, denote the vanishing ideal of $V$ by $I(V)$. If $A = (f_1,...,f_t)$ is a finitely generated ideal, define the Jacobian of A at a point $P \in k^n$ by $J_P(A) = \begin{pmatrix} \frac{\partial f_1}{\partial x_1}(P) & \dots &\frac{\partial f_1}{\partial x_n}(P)\\ \vdots & \dots & \vdots\\ \frac{\partial f_t}{\partial x_1}(P) & \dots &\frac{\partial f_t}{\partial x_n}(P) \end{pmatrix}$.
$\textbf{The contradiction I got:}$
In $\mathbb{C}[x,y,z,a,b,c]$, consider the ideals $A = \langle a^2+b^2+c^2,a(x+y+z),b(x+y+z),c(x+y+z)\rangle$ and $J = \langle a^2+b^2+c^2,x+y+z \rangle$.
Then, $A \subseteq J \implies V(J) \subseteq V(A)$, where V(J) denotes variety of J
$\implies \dim V(A) \geq \dim V(J) \geq 6-2 = 4 \hspace{1cm} [\because \dim V(f_1,...,f_r) \geq n-r$ in $k[x_1,...,x_n]$ and here $n=6,r=2$]
$\implies\forall P \in V(A)$ , rank $J_P(I(V(A)) \leq 6 - \dim V(A)\hspace{0.4cm}$ [A proof of this is given here ]
$\hspace{6.8cm}\leq 6-4=2$
Now obviously $A \subseteq I(V(A) \implies$ rank $J_P(A) \leq$ rank $J_P(I(V(A)) \hspace{0.5cm}\dots\dots(1)$ [I have given a proof of this at the end of this question]
$\implies$ rank $J_P(A) \leq 2 \hspace{1cm} \dots\dots(2)$
But if we take the point $P = (1,1,1,0,0,0) \in V(A)$ [here the coordinates are $x,y,z,a,b,c$-coordinates in given order], then $J_P(A) = \begin{pmatrix} \frac{\partial f_1}{\partial a}(P) & \frac{\partial f_1}{\partial b}(P) & \frac{\partial f_1}{\partial c}(P) &\frac{\partial f_1}{\partial x}(P) & \frac{\partial f_1}{\partial y}(P) & \frac{\partial f_1}{\partial z}(P)\\ \vdots & \dots & \dots & \dots & \dots & \vdots\\ \frac{\partial f_4}{\partial a}(P) & \frac{\partial f_4}{\partial b}(P) & \frac{\partial f_4}{\partial c}(P) &\frac{\partial f_4}{\partial x}(P) & \frac{\partial f_4}{\partial y}(P) & \frac{\partial f_4}{\partial z}(P) \end{pmatrix}$, where $f_1 = a^2+b^2+c^2, f_2 = a(x+y+z), f_3 = b(x+y+z), f_4 = c(x+y+z)$.
The only three non-zero entries in the above matrix are $\frac{\partial f_2}{\partial a}(P),\frac{\partial f_3}{\partial b}(P),\frac{\partial f_4}{\partial c}(P)$, which lie along a diagonal.
$\therefore$ rank $J_P(A) = 3$, which contradicts with $(2)$. Which step was wrong here?
$\textbf{Proof of (1):}$ We have to prove that for finitely generated ideals $A = \langle f_1,...,f_t\rangle$ and $B = \langle g_1,...,g_r\rangle$, if $A \subseteq B$, then rank $J_P(A) \leq$ rank $J_P(B)$ at any point $P \in V(B)$.
As $A \subseteq B$, for $1 \leq i \leq t$ express $f_i$ as $f_i = \sum_{j=1}^r g_jh_{ij}$
$\implies \frac{\partial f_i}{\partial x_k}(P) = \sum_{j=1}^r \left(\frac{\partial g_j}{\partial x_k}(P)h_{ij}(P) + g_j(P)\frac{\partial h_{ij}}{\partial x_k}(P)\right) = \sum_{j=1}^r \frac{\partial g_j}{\partial x_k}(P)h_{ij}(P)\hspace{0.2cm}$ [$\because g_j(P) = 0 \hspace{0.2cm} \forall P \in V(B)$]
$\implies$ A basis for row space of $J_P(B)$ gives a basis for row space of $J_P(A)$.
$\implies$ rank $J_P(A) \leq$ rank $J_P(B)$
For $A=(xz, yz)$ in $k[x, y, z]$ union of z-axis and the x-y plane.
$I(V(A))=(xz, yz)$.
$\dim V(A)=2$ and $n=3$.
But $\operatorname{rank} J_{((0, 0, 1)}(I(V(A))=2$ which is not $\leq 1=3-2=3-\dim V(A).$