By Vieta's: $p+q=k+3,\ pq=2k+1.$

$

(p-q)^{2}=(p+q)^{2}-4pq=(k+3)^{2}-4(2k+1)=k^{2}+6k+9-8k-4=k^{2}-2k+5.

$

Set $(p-q)^{2}=12$:

$

k^{2}-2k+5=12 ;\Longrightarrow; k^{2}-2k-7=0.

$

Discriminant $=4+28=32>0$, real distinct roots. Sum of roots $=2.$

Verification of root reality: For each such $k$, $(p-q)^2 = 12 > 0$, so $p, q$ are real and distinct. Both values of $k$ are admissible.

Why this is CAT-level: Mild but real. The standard Vieta route is direct, but the natural error is forgetting the $4pq$ term sign or confusing $(p-q)^2$ with $(p+q)^2$. The wrong options trap exactly those errors: option (A) $-2$ corresponds to summing roots of $k^2+2k-7$ (sign slip on the linear term).

Answer: (B) $2$.