Answer: \(k=\dfrac{212}{15}\).

We start with the main method. Differentiate to obtain the tangent gradient, then use the negative reciprocal for the normal gradient where needed.

The curve is

\(y=3(5x+6)^{1/2}.\)

At \(x=2\),

\(y=3(16)^{1/2}=12.\)

Differentiate:

\(\frac{dy}{dx}=3\cdot\frac12(5x+6)^{-1/2}\cdot5.\)

So

\(\frac{dy}{dx}=\frac{15}{2}(5x+6)^{-1/2}.\)

At \(x=2\),

\(\frac{dy}{dx}=\frac{15}{2}\cdot\frac14=\frac{15}{8}.\)

This is the gradient of the tangent, so the gradient of the normal is

\(-\frac{8}{15}.\)

The normal passes through \((2,12)\), so its equation is

\(y-12=-\frac{8}{15}(x-2).\)

The point \((-2,k)\) lies on this line. Substitute \(x=-2\):

\(k-12=-\frac{8}{15}(-4)=\frac{32}{15}.\)

Therefore

\(k=12+\frac{32}{15}=\frac{212}{15}.\)

This completes the solution and gives the required result.