At points of intersection,

\( kx^2-5x-6=3x-7k \).

So

\( kx^2-8x+7k-6=0 \).

For the line to intersect the curve, this equation must have real roots, so the discriminant is non-negative:

\( (-8)^2-4(k)(7k-6)\geq 0 \).

\( 64-28k^2+24k\geq 0 \).

\( 7k^2-6k-16\leq 0 \).

Factorise:

\( (7k+8)(k-2)\leq 0 \).

Answer: \( -\frac87\leq k\leq 2 \).