Given, P(x₁, y₁) = (1, 5)

Q(x₂, y₂) = (a, 1)

R(x₃, y₃) = (4, 11)

Since, the given points are collinear, The triangle formed by the giver points are to zero.

∴ Area of $\mathrm{△}\mathrm{PQR}=\frac{1}{2}\left|\left[{\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right]\right|=0$

$\Rightarrow \frac{1}{2}\left|\right[1(1-11)+\mathrm{a}(11-5)+4(5-1)\left]\right|=0$

$\Rightarrow \frac{1}{2}\left|\right[-10+16+6\mathrm{a}\left]\right|=0$

⇒ – 10 + 16 + 6a = 0

⇒ 6 + 6a = 0

⇒ 6(1 + a) = 0

⇒ a = – 1

∴ The value of a is – 1