Let the point, on the line passing through the points P(1, −2, 3) and Q(5, −4, 7), farther from the origin and at distance of 9 units from the point P, be $$(\alpha, \beta, \gamma)$$. Then $$\alpha^2 + \beta^2 + \gamma^2$$ is equal to:

We need to find the point on the line through P(1,-2,3) and Q(5,-4,7) that is 9 units from P and farther from the origin.

$$\vec{PQ} = Q - P = (4, -2, 4)$$

$$|\vec{PQ}| = \sqrt{16+4+16} = \sqrt{36} = 6$$

Unit vector: $$\hat{u} = \left(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right)$$

There are two such points (in both directions along the line):

$$P + 9\hat{u} = \left(1+6, -2-3, 3+6\right) = (7, -5, 9)$$

$$P - 9\hat{u} = \left(1-6, -2+3, 3-6\right) = (-5, 1, -3)$$

$$(7,-5,9)$$: distance$$^2 = 49+25+81 = 155$$

$$(-5,1,-3)$$: distance$$^2 = 25+1+9 = 35$$

$$(7,-5,9)$$ is farther.

$$7^2 + (-5)^2 + 9^2 = 49 + 25 + 81 = 155$$

The correct answer is Option 3: 155.