If $a,b,c\in\mathbb{R}$ such that $abc=p$ and $qa-b=0$, where $p,q$ are fixed positive number . Then the minimum distance of the point $(a,b,c)$ from the origin in $3$ dimension coordinate system
Options $\displaystyle (a)\; \sqrt{3}\bigg(\frac{p(q^2+1)}{2q}\bigg)^{\frac{1}{3}}$
$\displaystyle (b)\; \sqrt{3}\bigg(\frac{p(q^2+1)}{q}\bigg)^{\frac{1}{3}}$
$(c)\; \sqrt{3}(p)^{\frac{1}{3}}\;\;\;\;\;\; (d)\;\; \sqrt{2}(p/q)^{\frac{1}{2}}$
Try: Distance of $(a,b,c)$ from origin is $\sqrt{a^2+b^2+c^2}$
From $b=2a$ and $\displaystyle c=\frac{p}{ab}=\frac{p}{2a^2}$
Could some help me to solve it , thanks
Hint $b=qa$ and $c=p/qa^2$
By AM GM $$\frac {a^2+b^2+c^2}{3}\ge \sqrt[3] {a^2b^2c^2}=\sqrt[3] {a^2\cdot q^2a^2\cdot \frac {p^2}{q^2a^4}}=\sqrt[3] {p^2}$$
Hence answer is option C