Pairwise Products

$(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz \\ = \frac{(xy)(xz)}{yz} + \frac{(xy)(yz)}{xz} + \frac{(xz)(yz)}{xy} + 2\cdot (xy + xz + yz) \\ = 48 + 12 + 18\tfrac34 + 2\cdot 69 = 216\tfrac34 = \frac{867}4 = 3\cdot\frac{17^2}{2^2},$

so that

$x + y + z = \sqrt 3\cdot \frac{17}2$

so the answer is $3+17+2 = 22.$

$\begin{cases} xy = 24 & \Rightarrow y = \dfrac{24}{x} \\ zx = 30 & \Rightarrow z = \dfrac{30}{x} \\ yz = 15 = \dfrac{24\times 30}{x^2} & \Rightarrow x^2 = 48 \end{cases}$

$\Rightarrow x = 4\sqrt{3} \quad \Rightarrow y = \dfrac{24}{4\sqrt{3}} = 2\sqrt{3} \quad \Rightarrow z = \dfrac{30}{4\sqrt{3}} = \dfrac{5\sqrt{3}}{2}$

$\begin{aligned} x+y+z & = 4\sqrt{3} + 2\sqrt{3} + \dfrac{5\sqrt{3}}{2} = \dfrac{17\sqrt{3}}{2} \end{aligned}$

$\Rightarrow a+b+c =22.$

I like this question .Nice question Sir .
Given→ $xy = 24$ or $x=\dfrac{24}{y}$ .... $(1)$ $xz=30$ or $z=\dfrac{30}{x}$ ...... $(2)$
$yz=15$ or $y=\dfrac{15}{z}$ ....... $(3)$
Multiplying all three given conditions.
$xz×yz×zx=24×15×30=10800$
$x^2y^2z^2=10800$
$xyz=\sqrt{10800}$
Now adding $(1)$ , $(2)$ and $(3)$ .
$\dfrac{24}{y}+\dfrac{15}{z}+\dfrac{30}{x}$
$\dfrac{24zx+15xy+30yz}{xyz}$
$\dfrac{3(8zx+5xy+10yz)}{xyz}$
$\dfrac{3(8×30+5×24+10×15)}{\sqrt{10800}}$ $\dfrac{1530}{\sqrt{10800}}=\dfrac{17\sqrt{3}}{2}$
$\Rightarrow\dfrac{17\sqrt{3}}{2}$

$x=\frac{24}{y}=\frac{30}{z}$

$x^{2}=\frac{24 \times 30}{y \times z}=48$ since $y \times z = 15$

Hence $x = +4\sqrt{3}$ as we are told $x,y,z$ are positive.

Substitute our value for $x$ into the given equations.