Step 1: List what we know.
The object moves along the $x$-axis with uniform acceleration. At the start its velocity is $u = 12\ \text{cm s}^{-1}$ at position $x_0 = 3\ \text{cm}$, and after $t = 2\ \text{s}$ it sits at $x = -5\ \text{cm}$. We want the acceleration $a$.
Step 2: Choose the right equation of motion.
Since acceleration is constant and we know initial velocity, both positions and the time, the displacement equation is perfect: \[ x = x_0 + ut + \tfrac{1}{2}at^2 \]
Step 3: Substitute the numbers.
\[ -5 = 3 + (12)(2) + \tfrac{1}{2}a(2)^2 \]
Step 4: Simplify the known terms.
The term $(12)(2) = 24$ and $\tfrac{1}{2}a(4) = 2a$, so \[ -5 = 3 + 24 + 2a = 27 + 2a \]
Step 5: Solve for the acceleration.
Move $27$ across: \[ 2a = -5 - 27 = -32 \] \[ a = -16\ \text{cm s}^{-2} \] The negative sign tells us the acceleration points along $-\hat{i}$, which makes sense because the body actually ends up behind where it started.
Step 6: Write the answer as a vector.
Therefore the acceleration is directed along the negative $x$-axis with magnitude $16\ \text{cm s}^{-2}$, matching option (1). \[ \boxed{\vec{a} = (-16\ \text{cm s}^{-2})\hat{i}} \]