Let $\lambda_1, \lambda_2, \lambda_3, b\in \mathbb{R}$ with $\lambda_1\neq 0$.
Show that the linear equation $\lambda_1x_1+\lambda_2x_2+\lambda_3x_3=b$ has a set of solutions of the form $\{v\}+\text{span}(v_1, v_2)$, with $v, v_1, v_2\in \mathbb{R}^3$ and $v_1, v_2$ are linearly independent.
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Could you give me a hint how we could show that, since I have no idea? I don't really know how to start.
Note that the rank of the system is one (since $\lambda_1 \neq 0$) and then the null space has dimension $2$ ($=3-1$).
To find the solution note that
$$v=\left(\frac b{\lambda_1},0,0\right)$$
is a particular solution and
$$v_1=\left(\lambda_2,-\lambda_1,0\right) \quad v_2=\left(\lambda_3,0,-\lambda_1\right)$$
are two linearly independent solutions to $\lambda_1x_1+\lambda_2x_2+\lambda_3x_3=0$ therefore all the solutions for $(x_1,x_2,x_3)$ are given by
$$\{v\}+\text{span}(v_1, v_2)$$