Answer: Intercepts \(\left(-\frac13,0\right)\), \((5,0)\), \((0,5)\); \(k=\frac{64}{3}\).

First find the zeros of \(3x^2-14x-5\):

\(3x^2-14x-5=0.\)

Using the quadratic formula,

\(x=\frac{14\pm\sqrt{196+60}}{6} =\frac{14\pm16}{6}.\)

So

\(x=-\frac13\quad\text{or}\quad x=5.\)

Therefore the graph meets the \(x\)-axis at

\(\left(-\frac13,0\right)\quad\text{and}\quad (5,0).\)

When \(x=0\),

\(y=|-5|=5,\)

so the graph meets the \(y\)-axis at \((0,5)\).

The quadratic \(3x^2-14x-5\) is negative between its roots, so the modulus reflects that part above the \(x\)-axis.

The turning point of \(3x^2-14x-5\) occurs at

\(x=\frac{-(-14)}{2(3)}=\frac73.\)

At this value,

\(3\left(\frac73\right)^2-14\left(\frac73\right)-5 =-\frac{64}{3}.\)

After reflection by the modulus, the maximum point of the middle section has height \(\frac{64}{3}\).

The equation \(\left|3x^2-14x-5\right|=k\) has exactly 3 solutions when the horizontal line touches this maximum point and also intersects the two outer branches. Hence

\(\boxed{k=\frac{64}{3}}.\)