Given

$\rm 2^{x - 3} \cdot 2 \cdot a^{1 - x} = 2^{3x - 5} \cdot a^{x - 2}$

By the law of indices, $\rm b^{m - n} = \frac{b^{m}}{b^{n}}$, we get,

$\rm or, \frac{2^{x}}{2^{3}} \cdot 2 \cdot \frac{a^{1}}{a^{x}} = \frac{2^{3x}}{2^{5}} \cdot \frac{a^{x}}{a^{2}}$

$\rm or, \frac{ 2^{x} \cdot 2 \cdot a}{2^{3} \cdot a^{x}} = \frac{2^{3x} \cdot a^{x}}{2^{5} \cdot a^{2}}$

Placing all the terms with variable power on the left-hand side and the remaining terms on the right-hand side of the equation, we get,

$\rm or, \frac{2 ^{x}}{2^{3x} \cdot a^{x} \cdot a^{x}} = \frac{2^{3}}{2 \cdot a\cdot 2^{5} \cdot a^{2}}$

By using the law of indices, $\rm b^{m} \cdot b^{n} = b^{m + n}$, we get,

$\rm or, \frac{2^{x}}{2^{3x} \cdot a^{x + x}} = \frac{2^{3}}{2^{1 + 5} \cdot a^{1 + 2}}$

$\rm or, \frac{2^{x}}{2^{3x} \cdot a^{2x}} = \frac{2^{3}}{2^{6} \cdot a^{3}}$

By using the law of indices, $\rm \frac{b^{m}}{b^{n}} = b^{m - n}$, we get,

$\rm or, \frac{2^{x - 3x}}{a^{2x}} = \frac{2^{3 - 6}}{a^{3}}$

$\rm or, \frac{2^{-2x}}{a^{2x}} = \frac{2^{-3}}{a^{3}}$

By using the law of indices, $\rm a^{-m} = \frac{1}{a^{m}$, we get,

$\rm or, \frac{1}{2^{2x}} \cdot \frac{1}{a^{2x} = \frac{1}{2^{3}} \cdot \frac{1}{a^{3}}$

$\rm or, \frac{1}{2^{2x} \cdot a^{2x}} = \frac{1}{2^{3} \cdot a^{3}}$

By using the law of indices, $\rm b^{m} \cdot c^{m} = (bc)^{m}$, we get,

$\rm or, \frac{1}{ (2a)^{2x}} = \frac{1}{ (2a)^{3}}$

$\rm or, (2a)^{3} = (2a)^{2x}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm or, 3 = 2x$

Dividing both sides of the equation by 2 to free x, we get,

$\rm or, \frac{3}{2} = \frac{2}{2} \cdot x$

$\rm \therefore x = \frac{3}{2}$

Hence, the required value of x is $\rm \frac{3}{2}$.