The vector function expressed by $F=a_x\left(5y-k_{1}z\right)+a_y\left(3z+k_{2}x\right)+a_z\left(k_{3}y-4x\right)$ Represents a conservative field, where ax, ay, az are unit vectors along x, y and z directions, respectively. The values of constant k₁, k₂, k₃ are given by:

The correct option is: k₁ = 4, k₂ = 5, k₃ = 3.

To find the values of the constants k₁, k₂, and k₃, we use the definition of a conservative vector field.
A vector field $F$ is conservative if and only if its curl is equal to the zero vector:
$\nabla \times F=0$

Let the vector field be written in terms of its components as:
$F=F_x{a}_x+F_y{a}_y+F_z{a}_z$
From the given expression of the vector field:
$F_x=5y-k_{1}z$
$F_y=3z+k_{2}x$
$F_z=k_{3}y-4x$

The curl of $F$ in Cartesian coordinates is given by the determinant:
$\nabla \times F=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ F_x& F_y& F_z\end{array}\right|$

Expanding the determinant, we get:
$\nabla \times F=a_x(\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z})-a_y(\frac{\partial F_z}{\partial x}-\frac{\partial F_x}{\partial z})+a_z(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y})$

For the field to be conservative, each component of the curl must be equal to zero. This yields three independent equations:
1) For the $a_x$ component:
$\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z}=0\Rightarrow \frac{\partial F_z}{\partial y}=\frac{\partial F_y}{\partial z}$
2) For the $a_y$ component:
$\frac{\partial F_z}{\partial x}-\frac{\partial F_x}{\partial z}=0\Rightarrow \frac{\partial F_z}{\partial x}=\frac{\partial F_x}{\partial z}$
3) For the $a_z$ component:
$\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}=0\Rightarrow \frac{\partial F_y}{\partial x}=\frac{\partial F_x}{\partial y}$

Now, we compute the partial derivatives required for each equation:
- From Equation (1):
$\frac{\partial }{\partial y}(k_{3}y-4x)=k_3$
$\frac{\partial }{\partial z}(3z+k_{2}x)=3$
Equating them gives:
$k_3=3$

- From Equation (2):
$\frac{\partial }{\partial x}(k_{3}y-4x)=-4$
$\frac{\partial }{\partial z}(5y-k_{1}z)=-k_1$
Equating them gives:
$-4=-k_1\Rightarrow k_1=4$

- From Equation (3):
$\frac{\partial }{\partial x}(3z+k_{2}x)=k_2$
$\frac{\partial }{\partial y}(5y-k_{1}z)=5$
Equating them gives:
$k_2=5$

Combining these results, we get:
k₁ = 4, k₂ = 5, k₃ = 3