Answer: B

Solution:

Since $P(x)R(x)$ has degree $6$ and $P(Q(x))$ has degree $3\deg Q,$ we need $\deg Q = 2.$ A quadratic $Q$ is determined by the ordered triple $(Q(1), Q(2), Q(3)).$

At $x = 1, 2, 3$ the right side vanishes, so $P(Q(x)) = 0,$ forcing each of $Q(1), Q(2), Q(3)$ into $\{1, 2, 3\}.$ That gives $27$ triples.

Five of them give a polynomial of degree less than $2:$ the constants from $(1,1,1), (2,2,2), (3,3,3)$ and the linear $Q(x) = x$ from $(1,2,3)$ and $Q(x) = 4 - x$ from $(3,2,1).$ The other $27 - 5 = 22$ triples are non-collinear and yield genuine quadratics.

Thus, the correct answer is B.