Step-by-Step Solution:

1. Identify the Governing Principle: The problem states that two balls collide in free space with no external force acting on them. This means the total linear momentum of the system (the two balls) must be conserved.

2. Calculate the Initial Momentum: The initial momenta of the two balls are given as ${\overrightarrow p _1} = p\widehat i$ and ${\overrightarrow p _2} = - p\widehat i$. The total initial momentum of the system, $\overrightarrow P_{initial}$, is the vector sum of the individual momenta: ${\overrightarrow P_{initial}} = {\overrightarrow p _1} + {\overrightarrow p _2} = p\widehat i + ( - p\widehat i) = \overrightarrow 0$ The total initial momentum of the system is the zero vector.

3. Apply the Conservation of Momentum: According to the principle of conservation of linear momentum, the total final momentum, $\overrightarrow P_{final}$, must be equal to the total initial momentum. ${\overrightarrow P_{final}} = {\overrightarrow P_{initial}}$ ${\overrightarrow {p'} _1} + {\overrightarrow {p'} _2} = \overrightarrow 0$ This condition, ${\overrightarrow {p'} _1} + {\overrightarrow {p'} _2} = \overrightarrow 0$, must be satisfied for any physically possible (allowed) outcome of the collision. We will now check each option against this condition.

4. Analyze the Options: The question asks which option(s) are NOT ALLOWED for any non-zero value of the given parameters ($p,{a_1},{a_2},{b_1},{b_2},{c_1},{c_2}$). This means if an option requires one of its parameters to be zero to satisfy momentum conservation, it is not an allowed outcome under the problem's conditions.

   A: ${\overrightarrow {p'} _1} = {a_1}\widehat i + {b_1}\widehat j + {c_1}\widehat k; {\overrightarrow {p'} _2} = {a_2}\widehat i + {b_2}\widehat j$ Let's calculate the total final momentum: ${\overrightarrow P_{final}} = {\overrightarrow {p'} _1} + {\overrightarrow {p'} _2} = ({a_1} + {a_2})\widehat i + ({b_1} + {b_2})\widehat j + ({c_1})\widehat k$ For momentum to be conserved, ${\overrightarrow P_{final}}$ must be the zero vector. This requires each component to be zero: $a_1 + a_2 = 0$ $b_1 + b_2 = 0$ $c_1 = 0$ The condition $c_1=0$ must hold. However, the problem asks to consider cases with non-zero values for the parameters, including $c_1$. If $c_1 \neq 0$, the z-component of the final momentum is non-zero, and momentum is not conserved. Therefore, this option is not allowed for any non-zero value of $c_1$.

   B: ${\overrightarrow {p'} _1} = {c_1}\widehat k; {\overrightarrow {p'} _2} = {c_2}\widehat k$ The total final momentum is: ${\overrightarrow P_{final}} = {\overrightarrow {p'} _1} + {\overrightarrow {p'} _2} = ({c_1} + {c_2})\widehat k$ For momentum to be conserved, we need $c_1 + c_2 = 0$, which means $c_2 = -c_1$. It is possible to choose non-zero values for $c_1$ and $c_2$ that satisfy this condition (e.g., $c_1=5$ and $c_2=-5$). Since a valid scenario exists with non-zero parameters, this option is allowed.

   C: ${\overrightarrow {p'} _1} = {a_1}\widehat i + {b_1}\widehat j + {c_1}\widehat k; {\overrightarrow {p'} _2} = {a_2}\widehat i + {b_2}\widehat j - {c_1}\widehat k$ The total final momentum is: ${\overrightarrow P_{final}} = {\overrightarrow {p'} _1} + {\overrightarrow {p'} _2} = ({a_1} + {a_2})\widehat i + ({b_1} + {b_2})\widehat j + ({c_1} - {c_1})\widehat k$ ${\overrightarrow P_{final}} = ({a_1} + {a_2})\widehat i + ({b_1} + {b_2})\widehat j$ For momentum to be conserved, we need $a_1 + a_2 = 0$ and $b_1 + b_2 = 0$. We can find non-zero values for $a_1, a_2, b_1, b_2$ that satisfy these conditions (e.g., $a_1=1, a_2=-1, b_1=2, b_2=-2$). The parameter $c_1$ can be any non-zero value. Thus, this option is allowed.

   D: ${\overrightarrow {p'} _1} = {a_1}\widehat i + {b_1}\widehat j; {\overrightarrow {p'} _2} = {a_2}\widehat i + {b_1}\widehat j$ The total final momentum is: ${\overrightarrow P_{final}} = {\overrightarrow {p'} _1} + {\overrightarrow {p'} _2} = ({a_1} + {a_2})\widehat i + ({b_1} + {b_1})\widehat j$ ${\overrightarrow P_{final}} = ({a_1} + {a_2})\widehat i + 2{b_1}\widehat j$ For momentum to be conserved, each component must be zero: $a_1 + a_2 = 0$ $2b_1 = 0 \implies b_1 = 0$ This outcome is only possible if $b_1=0$. The problem asks to consider cases with non-zero values for the parameters. If $b_1 \neq 0$, the y-component of the final momentum is non-zero, and momentum is not conserved. Therefore, this option is not allowed for any non-zero value of $b_1$.

5. Conclusion: The options that are not allowed because they violate the conservation of linear momentum for non-zero values of certain parameters are A and D.