Let $$L_1, L_2$$ be the lines passing through the point $$P(0, 1)$$ and touching the parabola $$9x^2 + 12x + 18y - 14 = 0$$. Let $$Q$$ and $$R$$ be the points on the lines $$L_1$$ and $$L_2$$ such that the $$\triangle PQR$$ is an isosceles triangle with base $$QR$$. If the slopes of the lines $$QR$$ are $$m_1$$ and $$m_2$$, then $$16(m_1^2 + m_2^2)$$ is equal to _______

The parabola is $$9x^{2}+12x+18y-14=0$$.
Solve for $$y$$ to recognise its shape:

$$18y = -\,9x^{2}-12x+14 \;\;\Longrightarrow\;\; y = -\tfrac12x^{2}-\tfrac23x+\tfrac79$$.

This is a vertical parabola that opens downwards. A line through the fixed point $$P(0,1)$$ has the form

$$y-1 = m(x-0)\;\;\Longrightarrow\;\; y = mx+1$$ $$-(1)$$

For this line to be tangent to the parabola, simultaneous substitution of $$y$$ from $$(1)$$ into the parabola must give a quadratic in $$x$$ having zero discriminant.

Substituting $$y = mx+1$$ in $$9x^{2}+12x+18y-14=0$$ gives

$$9x^{2}+12x+18(mx+1)-14 = 0$$ $$\Longrightarrow\; 9x^{2} + (12+18m)x + 4 = 0$$ $$-(2)$$

For tangency, discriminant $$\Delta = 0$$:

$$\Delta = (12+18m)^{2} - 4\cdot9\cdot4 = 0$$ $$\Longrightarrow\; (12+18m)^{2} = 144$$ $$\Longrightarrow\; 12+18m = \pm12$$.

Solving,

Case 1: $$12+18m = 12 \;\Rightarrow\; 18m = 0 \;\Rightarrow\; m = 0$$.
Case 2: $$12+18m = -12 \;\Rightarrow\; 18m = -24 \;\Rightarrow\; m = -\tfrac43$$.

Hence the two tangents through $$P$$ are

$$L_1:\; y = 1 \quad(\text{slope }0)$$
$$L_2:\; y = -\tfrac43x + 1 \quad(\text{slope }-\tfrac43)$$.

Choose any point $$Q(x_q,1)$$ on $$L_1$$ and any point $$R(x_r,\, -\tfrac43x_r +1)$$ on $$L_2$$. Side lengths from $$P(0,1)$$ are

$$PQ = \sqrt{(x_q-0)^{2}+(1-1)^{2}} = |x_q|,$$
$$PR = \sqrt{(x_r-0)^{2}+(-\tfrac43x_r+1-1)^{2}} = \sqrt{x_r^{2} + (\tfrac43x_r)^{2}} = |x_r|\sqrt{1+\tfrac{16}{9}} = \tfrac53|x_r|.$$

The triangle $$PQR$$ is required to be isosceles with equal sides $$PQ = PR$$, so

$$|x_q| = \tfrac53|x_r|.$$

Two possibilities arise:

Case A: $$x_q = \tfrac53x_r\;\;(x_q\ \text{and}\ x_r\ \text{same sign}).$$ Slope of $$QR$$ is $$m = \frac{y_R - y_Q}{x_r - x_q} = \frac{-\tfrac43x_r +1 -1}{x_r - \tfrac53x_r} = \frac{-\tfrac43x_r}{-\tfrac23x_r} = 2.$$

Case B: $$x_q = -\,\tfrac53x_r\;\;(x_q\ \text{and}\ x_r\ \text{opposite signs}).$$ Now $$m = \frac{-\tfrac43x_r}{x_r + \tfrac53x_r} = \frac{-\tfrac43x_r}{\tfrac83x_r} = -\tfrac12.$$

Thus the two possible slopes of $$QR$$ are $$m_1 = 2$$ and $$m_2 = -\tfrac12$$.

Finally,

$$16\left(m_1^{2}+m_2^{2}\right) = 16\left(2^{2} + \left(-\tfrac12\right)^{2}\right) = 16\left(4 + \tfrac14\right) = 16\cdot\tfrac{17}{4} = 68.$$

Hence the required value is $$\boxed{68}$$.