Find $\min x^2+y^2+z^2$

If $ax+by+cz=k$, prove that the minimum value of $x^2+y^2+z^2$ is $\dfrac{k^2}{a^2+b^2+c^2}$.

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Since $$(x^2+y^2+z^2)(a^2+b^2+c^2)-(ax+by+cz)^2=(ay-bx)^2+(bz-cy)^2+(cx-az)^2\geq 0,$$ we have $$x^2+y^2+z^2\geq \frac{(ax+by+cz)^2}{a^2+b^2+c^2}=\frac{k^2}{a^2+b^2+c^2}.$$
Alternatively, we can use geometric arguments. The normal of $ax+by+cz=k$ is $\vec{n_1}=(a,b,c)$. The normal of $x^2+y^2+z^2$ is $\vec{n_2}=(2x,2y,2z)$. Applying Cauchy Schwartz inequality for $\vec{n_1}$ and $\vec{n_2}$: $$\vec{n_1}\cdot\vec{n_2}\leq \lVert \vec{n_1}\rVert \lVert \vec{n_2}\rVert,$$ we have $$2(ax+by+cz)\leq 2\sqrt{a^2+b^2+c^2} \sqrt{x^2+y^2+z^2},$$ which means $$x^2+y^2+z^2\geq \dfrac{k^2}{a^2+b^2+c^2}.$$