Consider the system of linear equations in x, y, z : x + 2y + tz =...

For a homogeneous system of linear equations to have infinitely many solutions, the determinant of its coefficient matrix must be zero.

The coefficient matrix is:
$\Delta = \begin{vmatrix} 1 & 2 & t \\ 6 & 1 & 5t \\ 3 & t^2 & f(t) \end{vmatrix}$

Expanding the determinant along the first row, we get:
$\Delta = 1(1 \cdot f(t) - 5t \cdot t^2) - 2(6 \cdot f(t) - 3 \cdot 5t) + t(6 \cdot t^2 - 3 \cdot 1)$

$\Delta = f(t) - 5t^3 - 12f(t) + 30t + 6t^3 - 3t$

$\Delta = -11f(t) + t^3 + 27t$

Since the system has infinitely many solutions for all $t \in \mathbb{R}$, we must have $\Delta = 0$ for all $t \in \mathbb{R}$.

$-11f(t) + t^3 + 27t = 0$

$f(t) = \dfrac{t^3 + 27t}{11}$

To determine the nature of the function $f(t)$, we find its derivative with respect to $t$:

$f'(t) = \dfrac{3t^2 + 27}{11}$

Since $t^2 \ge 0$ for all $t \in \mathbb{R}$, we have $3t^2 + 27 \ge 27 > 0$.

Therefore, $f'(t) > 0$ for all $t \in \mathbb{R}$, which implies that $f(t)$ is a strictly increasing function on $\mathbb{R}$.

Answer: is strictly increasing on $\mathbb{R}$