The focus of the parabola $$y^{2}=4x+16$$ is the centre of the circle C of radius 5 . If the values of $$\lambda$$, for which C passes through the point of intersection of the lines 3x − y = 0 and $$x + \lambda y = 4$$, are $$\lambda_{1}$$ and $$\lambda_{2},\lambda_{1} < \lambda_{2}$$, then $$12\lambda_{1}+29\lambda_{2}$$ is equal to

$$y^2 = 4x + 16 = 4(x + 4)$$

This is $$Y^2 = 4X$$ where $$X = x + 4$$. Focus is at X = 1, Y = 0, i.e., $$(x, y) = (-3, 0)$$.

Centre = (-3, 0), radius = 5.

$$(x+3)^2 + y^2 = 25$$

From 3x - y = 0: y = 3x.

Substituting: x + 3λx = 4, so $$x = \frac{4}{1+3\lambda}$$ and $$y = \frac{12}{1+3\lambda}$$

This point lies on the circle

$$\left(\frac{4}{1+3\lambda}+3\right)^2 + \left(\frac{12}{1+3\lambda}\right)^2 = 25$$

Let $$k = 1 + 3\lambda$$:

$$\left(\frac{4+3k}{k}\right)^2 + \left(\frac{12}{k}\right)^2 = 25$$

$$k = -\frac{5}{2}$$ or $$k = 4$$

$$1 + 3\lambda = -\frac{5}{2} \Rightarrow 3\lambda = -\frac{7}{2} \Rightarrow \lambda = -\frac{7}{6}$$

$$1 + 3\lambda = 4 \Rightarrow 3\lambda = 3 \Rightarrow \lambda = 1$$

So $$\lambda_1 = -\frac{7}{6}$$ and $$\lambda_2 = 1$$ (since $$\lambda_1 < \lambda_2$$).

$$12\lambda_1 + 29\lambda_2 = 12 \times (-\frac{7}{6}) + 29 \times 1 = -14 + 29 = 15$$

The answer is 15.