If the vector function \[\overrightarrow {\rm{F}} = {{\rm{\hat a}}_{\rm{x}}}\left( {3{\rm{y}} – {{\rm{k}}_1}{\rm{z}}} \right) + {{\rm{\hat a}}_{\rm{y}}}\left( {{{\rm{k}}_2}{\rm{x}} – 2{\rm{z}}} \right) – {{\rm{\hat a}}_{\rm{z}}}\left( {{{\rm{k}}_3}{\rm{y}} + {\rm{z}}} \right)\] is irrotational, then the values of the constants k1, k2 and k3 respectively, are A. 0.3, -2.5, 0.5 B. 0.0, 3.0, 2.0 C. 0.3, 0.33, 0.5 D. 4.0, 3.0, 2.0

A vector field is irrotational if its curl is zero. The curl of a vector field is defined as follows:

$$\nabla \times \overrightarrow {\rm{F}} = \left( {\frac{\partial }{{\partial {\rm{x}}}} \frac{\partial {\rm{F}}z}{{\partial {\rm{y}}}} – \frac{\partial {\rm{F}}_y}{{\partial {\rm{x}}}} \right){{\rm{\hat a}}{\rm{z}}} + \left( {\frac{\partial }{{\partial {\rm{y}}}} \frac{\partial {\rm{F}}x}{{\partial {\rm{z}}}} – \frac{\partial {\rm{F}}_z}{{\partial {\rm{y}}}} \right){{\rm{\hat a}}{\rm{x}}} + \left( {\frac{\partial }{{\partial {\rm{z}}}} \frac{\partial {\rm{F}}y}{{\partial {\rm{x}}}} – \frac{\partial {\rm{F}}_x}{{\partial {\rm{z}}}} \right){{\rm{\hat a}}{\rm{y}}}$$

Substituting the given vector field into the curl operator, we get:

$$\nabla \times \overrightarrow {\rm{F}} = \left( {\frac{\partial }{{\partial {\rm{x}}}} \frac{\partial }{{\partial {\rm{y}}}} \left( -{{\rm{k}}1}{\rm{z}} \right) – \frac{\partial }{{\partial {\rm{y}}}} \frac{\partial }{{\partial {\rm{x}}}} \left( {{\rm{k}}_2}{\rm{x}} – 2{\rm{z}} \right) \right){{\rm{\hat a}}{\rm{z}}} + \left( {\frac{\partial }{{\partial {\rm{y}}}} \frac{\partial }{{\partial {\rm{z}}}} \left( {{\rm{k}}3}{\rm{y}} + {\rm{z}} \right) – \frac{\partial }{{\partial {\rm{z}}}} \frac{\partial }{{\partial {\rm{y}}}} \left( 3{\rm{y}} – {{\rm{k}}_1}{\rm{z}} \right) \right){{\rm{\hat a}}{\rm{x}}} + \left( {\frac{\partial }{{\partial {\rm{z}}}} \frac{\partial }{{\partial {\rm{x}}}} \left( -2{\rm{z}} \right) – \frac{\partial }{{\partial {\rm{x}}}} \frac{\partial }{{\partial {\rm{z}}}} \left( {{\rm{k}}3}{\rm{y}} + {\rm{z}} \right) \right){{\rm{\hat a}}{\rm{y}}}$$

Simplifying, we get:

$$\nabla \times \overrightarrow {\rm{F}} = \left( { -{{\rm{k}}2} + 3{{\rm{k}}_3}} \right){{\rm{\hat a}}{\rm{x}}} + \left( { -{{\rm{k}}3}} \right){{\rm{\hat a}}{\rm{y}}} + \left( { -{{\rm{k}}1}} \right){{\rm{\hat a}}{\rm{z}}}$$

For the vector field to be irrotational, the curl must be zero. Therefore, we must have:

$${ -{{\rm{k}}_2} + 3{{\rm{k}}_3}} = 0$$
$${ -{{\rm{k}}_3}} = 0$$
$${ -{{\rm{k}}_1}} = 0$$

Solving for ${\rm{k}}_1$, ${\rm{k}}_2$, and ${\rm{k}}_3$, we get:

$${\rm{k}}_1 = 0.3$$
$${\rm{k}}_2 = 3.0$$
$${\rm{k}}_3 = 0.5$$

Therefore, the correct answer is $\boxed{{\rm{k}}_1 = 0.3, {\rm{k}}_2 = 3.0, {\rm{k}}_3 = 0.5}}$.