Match List I with List II

LIST I	LIST II
A. The solution of $3x + 7 > 12$ is	I. $[-1, \infty)$
B. The solution of $\frac{3x+5}{2} \geq 1$ is	II. $\left[\frac{17}{8}, \infty\right)$
C. The solution of $2x + 5 < 7x + 9$ is	III. $\left(\frac{5}{3}, \infty\right)$
D. The solution of $6x - 5 \geq -2x + 12$ is	IV. $\left(-\frac{4}{5}, \infty\right)$

Choose the correct answer from the options given below:

A matrix has 24 elements , which of the following cannot be the possible order of the matrix

the least non-negative remainder when $3^{17}$ is divided by 7 is :

The solution set of the inequation $\frac{7+5x}{4} \geq \frac{x}{3} - 10$, $x \in \mathbb{R}$ is :

If the matrix $\begin{bmatrix} a & -2 & 5b \\ 2 & 0 & -15 \\ 15 & 3c & 0 \end{bmatrix}$ is skew-symmetric, then the value of $a^2 + b^2 + c^2$ is:

If the matrix $A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$ satisfies the equation $A^T A = I_3$, then $x^2 + y^2 + z^2$ is :

If $A = \begin{bmatrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{bmatrix}$ and $B = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$, then AB is equals to :