1. Find the direction vector of the line through $P(1,-2,3)$ and $Q(5,-4,7)$.

   $\overrightarrow{PQ} = (5-1,\,-4-(-2),\,7-3) = (4,-2,4)$

   Its magnitude is $|\overrightarrow{PQ}| = \sqrt{4^2+(-2)^2+4^2} = \sqrt{16+4+16} = \sqrt{36}=6$

2. Find the point on this line at distance 9 from $P$.

   The unit vector along $\overrightarrow{PQ}$ is $\hat{u}=\left(\frac{4}{6},\frac{-2}{6},\frac{4}{6}\right)=\left(\frac23,-\frac13,\frac23\right)$

   A point at distance $9$ from $P$ on the line can be in either direction: $P \pm 9\hat{u}$

   So the two possible points are $P+9\hat{u}=(1,-2,3)+9\left(\frac23,-\frac13,\frac23\right)=(1+6,-2-3,3+6)=(7,-5,9)$

   and $P-9\hat{u}=(1,-2,3)-9\left(\frac23,-\frac13,\frac23\right)=(1-6,-2+3,3-6)=(-5,1,-3)$

3. Choose the point farther from the origin.

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

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

   Since $155>35$, the point farther from the origin is $(\alpha,\beta,\gamma)=(7,-5,9)$

4. Compute $\alpha^2+\beta^2+\gamma^2$.

   $\alpha^2+\beta^2+\gamma^2 = 7^2+(-5)^2+9^2 = 155$

5. Check options.

   • A: $150$ ❌
   • B: $155$ ✅
   • C: $160$ ❌
   • D: $165$ ❌

Therefore, the correct answer is B.