Space diagonal

let $L,H$ and $W$ be the length, height and width of the cuboid, respectively

Then,

$LH=8$ $\color{#3D99F6}\large \implies$ $L=\dfrac{8}{H}$ $\color{#D61F06}(1)$

$WL=12$ $\color{#3D99F6}\large \implies$ $L=\dfrac{12}{W}$ $\color{#D61F06}(2)$

$WH=6$ $\color{#D61F06}(3)$

Since $L=L$ we have

$\dfrac{8}{H}=\dfrac{12}{W}$ $\color{#3D99F6}\large \implies$ $W=1.5H$ $\color{#D61F06}(4)$

Substitute $\color{#D61F06}(4)$ in $\color{#D61F06}(3)$ . We have

$1.5H(H)=6$ $\color{#3D99F6}\large \implies$ $H^2=4$ $\color{#3D99F6}\large \implies$ $\boxed{H=2}$

It follows that, $L=\dfrac{8}{2}=$ $\boxed{4}$ and $W=\dfrac{6}{2}=$ $\boxed{3}$

So the diagonal of the base is $5$ (from $3-4-5$ special right triangle).

Finally, by pythagorean theorem , the length of the space diagonal, $d$ , is

$d=\sqrt{5^2+2^2}=\sqrt{25+4}=\sqrt{29}$

So the answer to the problem is $\color{plum}\large \boxed{29}$ .

Let the lengths of the width, depth and height of the cuboid and the space diagonal be $x$ , $y$ , $z$ and $w$ as shown in the figure above. Then we have:

$\begin{cases} zx = 8 & ...(1) \\ yz = 6 & ...(2) \\ xy = 12 &...(3) \end{cases}$

$\begin{aligned} (1)(2): \quad {\color{#3D99F6}xy}z^2 & = 8 \times 6 & \small \color{#3D99F6} (3): \ xy = 12 \\ {\color{#3D99F6}12}z^2 & = 48 \\ \implies z & = 2, \ x = 4, \ y = 3 \end{aligned}$

By Pythagorean theorem , the space diagonal $w^2 = x^2+y^2+z^2 = 4^2+3^2+2^2 = 29$ $\implies w = \sqrt{29}$ $\implies a = \boxed{29}$ .