\lambda)-2-3 \lambda=0$$

This is the same as the given equation $$5x - 11y + bz = 6a - 1$$.

Now, comparing the coefficients of the corresponding variables in both equations:

$$\frac{\lambda+1}{5} = \frac{2-\lambda}{-11} = \frac{a+\lambda}{b} = \frac{2+3\lambda}{6a-1}$$

Solving for $$\lambda$$:

$$-11\lambda -11 = 10 - 5\lambda$$

$$6\lambda = -21 \Rightarrow \lambda = -\frac{7}{2}$$

Now, substituting the value of $$\lambda$$ back into the equations:

$$\frac{2-\lambda}{-11} = \frac{2+3\lambda}{6a-1} \Rightarrow \frac{2+\frac{7}{2}}{-11} = \frac{2-\frac{21}{2}}{6a-1}$$

From this equation, we find the value of a :

$$6a - 1 = 17 \Rightarrow a = 3$$

Now, substituting the value of $$a$$ and $$\lambda$$ into the equation:

$$\frac{2-\lambda}{-11} = \frac{a+\lambda}{b} \Rightarrow -\frac{1}{2} = \frac{3 - \frac{7}{2}}{b}$$

$$\Rightarrow -\frac{b}{2} = -\frac{1}{2} \Rightarrow b = 1$$

Therefore, the point $$(a, -c, c) \equiv (3, -c, c)$$.

The given distance is $$\frac{2}{\sqrt{a}} = \frac{2}{\sqrt{3}}$$.

The plane is: $$5x - 11y + z = 17$$.

Now, let's find the distance:

$$\left|\frac{15 + 11c + c - 17}{\sqrt{147}}\right| = \frac{2}{\sqrt{3}}$$

$$\Rightarrow c = -1, \frac{4}{3}$$

Since $$c \in \mathbb{Z}$$, we have $$c = -1$$.

Therefore, $$\frac{a+b}{c} = \frac{3+1}{-1} = -4$$.

3d Geometry

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