If A, B, and C are three singular matrices given by

The correct answer is 45.

A square matrix is said to be singular if its determinant is equal to zero. We are given three singular matrices: $A$, $B$, and $C$. We can find the values of the variables $a$, $b$, and $c$ step-by-step by setting their determinants to zero.

Step 1: Finding $a$ from matrix $A$
Based on the first image, the matrix $A$ is:
$$A=\left[\begin{array}{cc}a& 4\\ 3& 2\end{array}\right]$$
Since $A$ is a singular matrix, its determinant must be zero:
$$det\left(A\right)=(a\times 2)-(4\times 3)=0$$
$$2a-12=0$$
$$2a=12\Rightarrow a=6$$

Step 2: Finding $b$ from matrix $B$
Based on the second image, the matrix $B$ is:
$$B=\left[\begin{array}{cc}3b& 5\\ a& 2\end{array}\right]$$
Since $B$ is a singular matrix, its determinant must be zero:
$$det\left(B\right)=(3b\times 2)-(5\times a)=0$$
$$6b-5a=0$$
Substituting $a=6$ into the equation:
$$6b-5\left(6\right)=0$$
$$6b-30=0$$
$$6b=30\Rightarrow b=5$$

Step 3: Finding $c$ from matrix $C$
Based on the second image, the matrix $C$ is:
$$C=\left[\begin{array}{cc}a+b+c& c+1\\ a+c& c\end{array}\right]$$
Since $C$ is a singular matrix, its determinant must be zero:
$$det\left(C\right)=(a+b+c)\left(c\right)-(c+1)(a+c)=0$$
Expanding both terms of the expression:
$$(ac+bc+c^2)-(ac+c^2+a+c)=0$$
$$ac+bc+c^2-ac-c^2-a-c=0$$
Canceling the common terms $ac$ and $c^2$, we obtain:
$$bc-a-c=0$$
Substituting the values $a=6$ and $b=5$:
$$5c-6-c=0$$
$$4c=6$$
$$c=\frac{6}{4}=1.5$$

Step 4: Finding the value of $abc$
Using the values $a=6$, $b=5$, and $c=1.5$:
$$abc=6\times 5\times 1.5=45$$