(a)

Motion of the ball is as shown below.

We can combine all three portions one after other to get a perf ect parabola obtained by throwing ball from a point on the ground. It is as shown below.

Now, for the standard parabolic motion, equation of trajectory can be written as

$y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}$

Substituting $R=\frac{u^2 \sin 2 \theta}{g}$ in above equation, we have

$Y=X \tan \theta \cdot\left(1-\frac{X}{R}\right)$

When $y=14 \,m , . Y=x$ and $R=12 \,m$, so

We have $14=x \tan \theta \cdot\left(1-\frac{x}{12}\right) \quad \dots(i)$

Again, when $y=3 \,m$ and $X=6+x$, so We have

$3=(6+x) \tan \theta \cdot\left(1-\frac{6+x}{12}\right) \ldots(i)$

Dividing Eq. $(i)$ by Eq.$(ii)$, we get

$\frac{14}{3}=\frac{x(12-x)}{(6+x)(6-x)}$

$\Rightarrow 7\left(36-x^2\right)=15\left(12 x-x^2\right)$

$\Rightarrow \quad 2 x^2-45 x+63=0$

$\Rightarrow x^2-\left(\frac{3}{2}+21\right) x+\left(\frac{3}{2} \times 21\right)=0$

$\Rightarrow \quad x^2-\frac{3}{2} x-21 x+\frac{3}{2} \times 21=0$

$\Rightarrow \quad x\left(x-\frac{3}{2}\right)-21\left(x-\frac{3}{2}\right)=0$

$\Rightarrow \quad\left(x-\frac{3}{2}\right)(x-21)=0$

As, $x$ must be less than $6 \,m$.

So, $x=\frac{3}{2}=15 \,m$.