Basis for $v \in V=\mathbb Q^3$ such that $x_1\neq x_2$, in which coordinates are $(1,0,0)$

Question

Given $V=\mathbb Q^3$, and a vector $v=(x_1,x_2,x_3) \in V$ such that $x_1\neq x_2$. Is there any basis $B$ in $V$ in which the coordinates of the vector $v$ are $(1,0,0)$?

a) No, because $v$ could be the null vector.

b) We can't answer the question.

c) Yes, because $v$ is not the null vector.

d) It depends on the value of $x_3$.

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My attempt:

I don't know where to start. I'm lost, and I need at least a hint.

Answer 1

Hint: Asking if the coordinates of $v$ are $(1,0,0)$ is the same as asking if there is a basis for $V$ consisting of $v$ together with two other vectors $w$ and $x$. Since $x_1\ne x_2$, you know that $v$ is nonzero (right?). What do you know about extending a set of vectors to a basis?