Step 1: Approach
Use the determinant form of the plane through three points: two on the line and Q.
Step 2: Points
$P(1,-3,1)$ and $P'=P+\vec d=(3,-5,2)$ are on the line, and $Q(1,0,-1)$ is given.
Step 3: Plane
The plane through $Q$ with vectors $\overrightarrow{QP}=(0,-3,2)$ and $\vec d=(2,-2,1)$ has normal $\overrightarrow{QP}\times\vec d=(-3\cdot1-2(-2),\ 2\cdot2-0,\ 0-(-3)(2))=(1,4,6)$.
Step 4: Distance
Equation $x+4y+6z+5=0$, distance $\dfrac{5}{\sqrt{53}}$, so $p=5$. Option (C).
Final Answer:
The plane is x + 4y + 6z + 5 = 0, which is 5 over root 53 from the origin, so p = 5, option (C).
\[ \boxed{5} \]