The volume of the tetrahedron whose vertices are A$(-1, 2, 3)$, B$(3, -2, 1)$, C$(p, 1, 3)$, D$(-1, -2, 4)$ is $\dfrac{16}{3}$ cubic units then the value of p is

The volume of a tetrahedron with vertices $A, B, C, D$ is given by $V = \dfrac{1}{6} |[\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}]|$.

The vectors representing the edges from $A$ are:
$\overrightarrow{AB} = (3 - (-1))\hat{i} + (-2 - 2)\hat{j} + (1 - 3)\hat{k} = 4\hat{i} - 4\hat{j} - 2\hat{k}$
$\overrightarrow{AC} = (p - (-1))\hat{i} + (1 - 2)\hat{j} + (3 - 3)\hat{k} = (p+1)\hat{i} - \hat{j}$
$\overrightarrow{AD} = (-1 - (-1))\hat{i} + (-2 - 2)\hat{j} + (4 - 3)\hat{k} = -4\hat{j} + \hat{k}$

The scalar triple product is:
$[\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}] = \begin{vmatrix} 4 & -4 & -2 \\ p+1 & -1 & 0 \\ 0 & -4 & 1 \end{vmatrix}$

Expanding the determinant:
$= 4(-1 - 0) + 4(p+1 - 0) - 2(-4(p+1) - 0)$
$= -4 + 4p + 4 + 8p + 8$
$= 12p + 8$

Given $V = \dfrac{16}{3}$, we have:
$\dfrac{1}{6} |12p + 8| = \dfrac{16}{3}$

$|12p + 8| = 32$

$12p + 8 = \pm 32$

Taking the positive sign:
$12p + 8 = 32 \Rightarrow 12p = 24 \Rightarrow p = 2$

Taking the negative sign:
$12p + 8 = -32 \Rightarrow 12p = -40 \Rightarrow p = \dfrac{-10}{3}$

Since $p = \dfrac{-10}{3}$ is present in the options, it is the required value.

Answer: $\dfrac{-10}{3}$