Step 1: Apply column operations instead. Let the columns be $C_1=(a+pd,\,p,\,d)$, $C_2=(a+qd,\,q,\,d)$, $C_3=(a+rd,\,r,\,d)$.
Step 2: Replace $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$ (this does not change the determinant's value): $$C_2 - C_1 = \big((q-p)d,\ q-p,\ 0\big), \qquad C_3-C_1 = \big((r-p)d,\ r-p,\ 0\big)$$
Step 3: The determinant now reads $$\begin{vmatrix} a+pd & (q-p)d & (r-p)d \\ p & q-p & r-p \\ d & 0 & 0 \end{vmatrix}$$
Step 4: Expand along the third row, which has only one nonzero entry, $d$, in the first column. The determinant equals $d$ times its cofactor: $$d \cdot \begin{vmatrix} (q-p)d & (r-p)d \\ q-p & r-p \end{vmatrix}$$
Step 5: Evaluate this $2\times 2$ determinant: $$(q-p)d\cdot(r-p) - (r-p)d\cdot(q-p) = d(q-p)(r-p) - d(r-p)(q-p) = 0$$ since both terms are identical.
Step 6: So the whole expression is $d \cdot 0 = 0$.
\[\boxed{0}\]