Ta có: \({\log _a}b = 2{\log _b}c \Leftrightarrow {\log _a}b.{\log _b}c = 2\log _b^2c \Leftrightarrow {\log _a}c = 2\log _b^2c\)

Ta có: \({\log _a}b = 4{\log _c}a \Leftrightarrow {\log _a}b.{\log _c}a = 4\log _c^2a \Leftrightarrow {\log _c}b = 4\log _c^2a\).

Suy ra \({\log _c}b.{\log _a}c = 8\log _c^2a.\log _b^2c \Leftrightarrow {\log _a}b = 8\log _b^2a\)

\( \Leftrightarrow {\log _a}b = \frac{8}{{\log _a^2b}} \Leftrightarrow \log _a^3b = 8 \Leftrightarrow {\log _a}b = 2 \Leftrightarrow b = {a^2}\).

Mặt khác: \({\log _a}b = 2{\log _b}c \Leftrightarrow {\log _a}{a^2} = 2{\log _b}c \Leftrightarrow {\log _b}c = 1 \Leftrightarrow b = c\).

Theo giả thiết: \(a + 2b + 3c = 48 \Leftrightarrow a + 2{a^2} + 3{a^2} = 48 \Leftrightarrow 5{a^2} + a - 48 = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{a = 3}\\{a =  - \frac{5}}\end{array}} \right.\).

Do \(a > 0\) nên \(a = 3\). Với \(a = 3 \Rightarrow c = 3 \Rightarrow b = 9\). Vậy \(a + b + c = 15\). Đáp án: 15.