Answer :

Solution :

Given that,

$\overrightarrow{F}={\hat{a}}_x\left(5y-k_{1}z\right)+{\hat{a}}_y\left(3z+k_{2}x\right)+{\hat{a}}_z\left(k_{3}y-4x\right)$

is a conservative field.

$\left|\begin{array}{ccc}{\hat{a}}_x& {\hat{a}}_y& {\hat{a}}_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \left(5y-k_{1}z\right)& \left(3z+k_{2}x\right)& \left(k_{3}y-4x\right)\end{array}\right|=0$

k₃ – 3 = 0

⇒ k₃ = 3 -4 + k₁ = 0

⇒ k₁ = 4 k₂ – 5 = 0

⇒ k₂ = 5

The required values are: k₁ = 4, k₂ = 5, k₃ = 3