How would I find the (a,b) that satisfies that the tangent plane to $f(x,y) = (x^2) + 2xy + (y^2)$ passes through the point $(2,1,0)$ ?
I know that $f(x)= 2x + 2y$, and $F(y): 2x + 2y$.
Therefore $L(a,b): (2a+2b) (x-a) + (2a + 2b) (y-b) + f(a,b)$
But how can I find $(a,b)$?
The equation of the tangent plane to $f$ at $(a,b,f(a,b))$ is given by $$\pi_{(a,b)}:\quad z=f(a,b)+(x-a)f_x(a,b)+(y-b)f_y(a,b).$$ Given you want to find the pairs $(a,b)$ such that $\pi_{(a,b)}$ passes through the point $P=(2,1,0)$, the equation must be satisified when you change $(x,y,z)$ by $P$.
Hence, since you already found the equation of the tangent plane, you get $$(2a+2b)(2-a)+(2a + 2b)(1-b)+(a+b)^2=0$$ $$(a+b)(4-2a)+(a+b)(2-2b)+(a+b)^2=0$$ $$(a+b)(6-a-b)=0 \tag{1}$$ and any point $(a,b)$ that satisfies $(1)$ has the required condition.