Let A = bmatrix 1 & 3 & -1 2 & 1 & α 0 & 1 & -1 bmatrix be a...

Since $A$ is a singular matrix, $|A| = 0$.

$|A| = 1(-1 - \alpha) - 3(-2 - 0) - 1(2 - 0) = 0$

$-1 - \alpha + 6 - 2 = 0 \Rightarrow \alpha = 3$

The function is $f(x) = \int\limits_0^x (t^2 + 2t + 3)\,dt = \dfrac{x^3}{3} + x^2 + 3x$

Differentiating with respect to $x$, we get $f'(x) = x^2 + 2x + 3 = (x+1)^2 + 2 > 0$ for all $x$.

Thus, $f(x)$ is a strictly increasing function on the interval $[1, 3]$.

The maximum value $M$ occurs at $x = 3$:

$M = f(3) = \dfrac{3^3}{3} + 3^2 + 3(3) = 9 + 9 + 9 = 27$

The minimum value $m$ occurs at $x = 1$:

$m = f(1) = \dfrac{1^3}{3} + 1^2 + 3(1) = \dfrac{13}{3}$

We need to find $3(M - m)$:

$3(M - m) = 3\left(27 - \dfrac{13}{3}\right) = 81 - 13 = 68$

Answer: $68$