Behavior of Function for System of Linear Equations

To determine the behavior of the differentiable function $f(t)$, we analyze the system of homogeneous linear equations: $x + 2y + tz = 0$ $6x + y + 5tz = 0$ $3x + t^2y + f(t)z = 0$

For a homogeneous system of equations to have non-trivial (and hence infinitely many) solutions, the determinant of the coefficient matrix must be equal to zero. Since it is given that the system has infinitely many solutions for all $t \in \mathbb{R}$, the determinant must be identically zero for all $t \in \mathbb{R}$.

The coefficient matrix $A$ is given by: $A = \begin{pmatrix} 1 & 2 & t \\ 6 & 1 & 5t \\ 3 & t^2 & f(t) \end{pmatrix}$

We calculate the determinant $\det(A)$: $\det(A) = \begin{vmatrix} 1 & 2 & t \\ 6 & 1 & 5t \\ 3 & t^2 & f(t) \end{vmatrix}$

Expanding along the first row: $\det(A) = 1 \cdot (1 \cdot f(t) - 5t \cdot t^2) - 2 \cdot (6 \cdot f(t) - 5t \cdot 3) + t \cdot (6 \cdot t^2 - 1 \cdot 3)$ $\det(A) = (f(t) - 5t^3) - 2(6f(t) - 15t) + t(6t^2 - 3)$ $\det(A) = f(t) - 5t^3 - 12f(t) + 30t + 6t^3 - 3t$ $\det(A) = -11f(t) + t^3 + 27t$

Setting $\det(A) = 0$ for all $t \in \mathbb{R}$: $-11f(t) + t^3 + 27t = 0$ $f(t) = \frac{t^3 + 27t}{11}$

Now, to determine the nature of the function $f(t)$, we take its derivative with respect to $t$: $f'(t) = \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{t^3 + 27t}{11}\right) = \frac{3t^2 + 27}{11}$

Since $t^2 \ge 0$ for all $t \in \mathbb{R}$: $3t^2 + 27 \ge 27 > 0 \quad \forall t \in \mathbb{R}$

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

Hence, the correct option is B.