I'm struggling with a simple question of the homework.
Consider a set P s.t $$P = \{x \in R^{3} : x_1 + 2x_2 + 3x_3 = 1\}$$
Then it was asked to prove that this set is an affine set. I heard that one way to answer this question is to select two points $s_1$, $s_2$ s.t $s_1,s_2 \in P$ and show that $x(\theta) = \theta*x_1 + (1 - \theta) * x_2 \ \forall \ \theta$, but still I don't know how to start. For some way, I have a thought that induction is needed, or that I am missing some concepts.
There's an issue here about the meaning of "select two points" - to verify the definition (given in the comments), it doesn't suffice to pick two particular points $p$ and $q$ in $P$ and verify that $(1-t)p + tq$. We have to verify this for all pairs of points. But often people will say "select two points $p,q\in P$" when they want to reason about arbitrary points.
Anyway, this comes down to a simple computation verifying the definition. Let $p,q\in P$. Let's write $p = (a_1,a_2,a_3)$ and $q = (b_1,b_2,b_3)$. We know $a_1+2a_2+3a_3 = 1$ and $b_1 + 2b_2 + 3b_3 = 1$.
Now we need to show $(1-t)p + tq \in P$ for any $t\in \mathbb{R}$. Let's call this point $r = (c_1,c_2,c_3)$. Let's write out the coordinates of this point: \begin{align*} r &= (1-t)p + tq\\ &= (1-t)(a_1,a_2,a_3) + t(b_1,b_2,b_3)\\ &= ((1-t)a_1, (1-t)a_2, (1-t)a_3)) + (tb_1,tb_2,tb_3)\\ &= ((1-t)a_1 + tb_1,(1-t)a_2 + tb_2,(1-t)a_3 + tb_3) \end{align*}
Now let's check if it's in $P$: \begin{align*} c_1 + 2c_2 + 3c_3 &= ((1-t)a_1 + tb_1) + 2((1-t)a_2 + tb_2) + 3((1-t)a_3 + tb_3)\\ &= (1-t)(a_1 + 2a_2 + 3a_3) + t(b_1 + 2b_2 + 3b_3)\\ &= (1-t)(1) + t(1) \\ &= 1 \end{align*} So it is!
You can easily abstract from this argument to show that for any $a,b,c,d\in \mathbb{R}$, the set $$P = \{(x,y,z)\in \mathbb{R}^3\mid ax + by + cz = d\}$$ is affine.