Directions (46-47): Read the following quadric equation carefully and answer the questions given below.

Equation 1: $a{x}^2+bx+c=0$
Equation 2: $d{y}^2+ey+c=0$

Note:
(i) c is a single digit prime number greater than 2
(ii) d and b are two digits prime number less than 20
(iii) d is greater than 11
(iv) b is greater than d
(v) Smallest roots of both the equation are same
(vi) No root is irrational
(vii) e = b + 1
(viii) 3c is greater than b

Find the value of ‘a’ .

To find the value of $a$, let us analyze the given conditions step-by-step:

Step 1: Determine the value of $c$
According to Note (i), $c$ is a single-digit prime number greater than 2.
The single-digit prime numbers are 2, 3, 5, and 7. Since $c>2$, $c$ must be one of $\{3,5,7\}$.
According to Note (viii), $3c>b$, or $c>\frac{b}{3}$.

Step 2: Determine the values of $d$ and $b$
According to Note (ii), $d$ and $b$ are two-digit prime numbers less than 20.
The two-digit prime numbers less than 20 are 11, 13, 17, and 19.
According to Note (iii), $d>11$. Thus, $d$ must be one of $\{13,17,19\}$.
According to Note (iv), $b>d$. This gives us the following possible pairs for $(d,b)$:
1. If $d=13$, then $b$ can be 17 or 19.
2. If $d=17$, then $b$ can only be 19.
3. If $d=19$, there is no prime less than 20 greater than $d$.

Now, let us use the condition $3c>b$:
Since the minimum possible value for $b$ is 17, we must have $3c>17\Rightarrow c>5.67$.
Out of the candidates $\{3,5,7\}$, only $c=7$ satisfies this inequality.
Thus, we must have $c=7$.

Step 3: Analyze Equation 2 to find $d$, $b$, and the roots
Equation 2 is given by:
$d{y}^2+ey+c=0$
According to Note (vii), $e=b+1$.
Since no root is irrational (Note vi), the discriminant of Equation 2, $D=e^2-4dc$, must be a perfect square.

Let us test the possible configurations for $(d,b)$ with $c=7$:
- Case A: $d=13$, $b=17$:
Here, $e=17+1=18$.
The discriminant is $D={18}^2-4\left(13\right)\left(7\right)=324-364=-40$ (negative, hence imaginary roots). This case is invalid.

- Case B: $d=17$, $b=19$:
Here, $e=19+1=20$.
The discriminant is $D={20}^2-4\left(17\right)\left(7\right)=400-476=-76$ (negative, hence imaginary roots). This case is invalid.

- Case C: $d=13$, $b=19$:
Here, $e=19+1=20$.
The discriminant is $D={20}^2-4\left(13\right)\left(7\right)=400-364=36$.
Since 36 is a perfect square ($6^2$), this configuration is valid. The roots of Equation 2 are:
$y=\frac{-20\pm \sqrt{36}}{2\left(13\right)}=\frac{-20\pm 6}{26}$
The two roots are:
$y_1=\frac{-20-6}{26}=-1$
$y_2=\frac{-20+6}{26}=-\frac{7}{13}$
The smallest root of Equation 2 is $-1$.

Step 4: Find the value of $a$ using Equation 1
According to Note (v), the smallest roots of both equations are the same. Thus, the smallest root of Equation 1 is also $-1$.
Substituting $x=-1$ into Equation 1:
$a{x}^2+bx+c=0$
$a{(-1)}^2+b(-1)+c=0$
$a-b+c=0$
Substitute the known values $b=19$ and $c=7$:
$a-19+7=0$
$a-12=0\Rightarrow a=12$.

The correct option corresponding to this value is 12.