Let the equation of the plane P containing the line $$x + 10 = \frac{8-y}{2} = z$$ be $$ax + by + 3z = 2(a+b)$$ and the distance of the plane P from the point $$(1, 27, 7)$$ be $$c$$. Then $$a^2 + b^2 + c^2$$ is equal to

The given equation of the line can be written in standard symmetric form as:

$$\frac{x + 10}{1} = \frac{y - 8}{-2} = \frac{z - 0}{1}$$

From this line, we can identify a passing point $$P_0(-10, 8, 0)$$ and its direction vector $$\vec{d} = \hat{i} - 2\hat{j} + \hat{k}$$.

Since the plane $$ax + by + 3z = 2(a+b)$$ contains this line, the point $$P_0$$ must lie on the plane:

$$a(-10) + b(8) + 3(0) = 2(a+b)$$

$$-10a + 8b = 2a + 2b$$

$$-12a + 6b = 0 \implies b = 2a$$

Additionally, the direction vector of the line must be perpendicular to the normal vector of the plane $$\vec{n} = a\hat{i} + b\hat{j} + 3\hat{k}$$:

$$\vec{d} \cdot \vec{n} = 0$$

$$1(a) - 2(b) + 1(3) = 0$$

$$a - 2b + 3 = 0$$

Substituting $$b = 2a$$ into this orthogonality condition:

$$a - 2(2a) + 3 = 0$$

$$-3a + 3 = 0 \implies a = 1$$

$$b = 2(1) = 2$$

Substituting $$a = 1$$ and $$b = 2$$ gives the equation of the plane $$P$$:

$$x + 2y + 3z = 2(1+2) \implies x + 2y + 3z - 6 = 0$$

The distance $$c$$ from the point $$(1, 27, 7)$$ to the plane is given by the perpendicular distance formula:

$$c = \frac{|1(1) + 2(27) + 3(7) - 6|}{\sqrt{1^2 + 2^2 + 3^2}}$$

$$c = \frac{|1 + 54 + 21 - 6|}{\sqrt{14}} = \frac{70}{\sqrt{14}} = 5\sqrt{14}$$

Squaring the distance yields:

$$c^2 = (5\sqrt{14})^2 = 25 \times 14 = 350$$

Evaluating the final expression $$a^2 + b^2 + c^2$$:

$$a^2 + b^2 + c^2 = 1^2 + 2^2 + 350 = 1 + 4 + 350 = 355$$

Conclusion:

The value of $$a^2 + b^2 + c^2$$ is equal to 355.