Let the line $$\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z-2}{2}$$ lies in the plane $$x + 3y - \alpha z + \beta = 0$$. Then $$(\alpha, \beta)$$ equals

1. Analyze the Direction Vector Conditions

If a straight line lies entirely inside a flat plane, two conditions must be satisfied:

The direction vector of the line must be perpendicular to the normal vector of the plane.

Any specific point situated on the line must satisfy the equation of the plane.

From the symmetrical form of the line equation, extract its direction vector:

$$\vec{d} = 3\hat{i} - 5\hat{j} + 2\hat{k}$$

From the general Cartesian equation of the plane, extract its normal vector:

$$\vec{n} = 1\hat{i} + 3\hat{j} - \alpha\hat{k}$$

2. Apply the Perpendicular Vector Condition

Since the line is perpendicular to the normal vector, their vector dot product must be equal to zero:

$$(3)(1) + (-5)(3) + (2)(-\alpha) = 0$$

$$3 - 15 - 2\alpha = 0$$

$$-12 - 2\alpha = 0$$

$$2\alpha = -12 \implies \alpha = -6$$

3. Apply the Point Alignment Condition

Identify a known coordinate point P passing through the given line equation:

$$P = (2, 1, 2)$$

Because the line lies completely inside the plane, this point $$P(2, 1, 2)$$ must satisfy the plane equation. Substitute these coordinate values along with our calculated value of $$\alpha = -6$$ into the plane relation:

$$(2) + 3(1) - (-6)(2) + \beta = 0$$

$$2 + 3 + 12 + \beta = 0$$

$$17 + \beta = 0 \implies \beta = -17$$

4. Combine the Parameters

Grouping the two solved parameter values together gives the ordered coordinate pair:

$$(\alpha, \beta) = (-6, -17)$$