To solve this, we need to determine the kernel of the linear transformation $T : \mathbb{R}^3 \to \mathbb{R}^2$ given by:

[ T[x, y, z] = [2x - 5y + z, x + 3z] ]

The kernel of a linear transformation consists of all vectors $[x, y, z]$ in $\mathbb{R}^3$ such that $T[x, y, z] = [0, 0]$. This means we need to solve the following system of equations:

1. $2x - 5y + z = 0$
2. $x + 3z = 0$

Let's proceed to solve this system:

From the second equation, we can express $x$ as: $x = -3z$

Substitute this into the first equation: $2(-3z) - 5y + z = 0$ $-6z - 5y + z = 0$ $-5z - 5y = 0$ $y = -z$

Thus, we have $x = -3z$ and $y = -z$, and we can write the general solution as: [ [x, y, z] = z[-3, -1, 1] ]

Therefore, the kernel of $T$ is spanned by the vector $[-3, -1, 1]$, and we can express the kernel as: $\ker(T) = \{(-3, -1, 1)t \mid t \in \mathbb{R}\}$

Thus, the correct answer is: $\boxed{\ker(T) = \{(-3, -1, 1)t \mid t \in \mathbb{R}\}}$

Would you like further explanation on this or have any more questions related to it?

Here are a few related questions:

1. How do you determine the kernel of a linear transformation?
2. What is the rank-nullity theorem, and how does it relate to the kernel?
3. How do you interpret the kernel geometrically?
4. What is the difference between the kernel and image of a linear transformation?
5. How do you find the image of a linear transformation?

Tip: To solve systems of equations involving linear transformations, always begin by setting the transformation equal to the zero vector and solving the system for the unknowns.