If $A = \begin{bmatrix} 3 & -2 \\ 4 & 2 \end{bmatrix}$, then $A^2 - 5A + 14I = 0$. Which of the following is equivalent to $A^2$?

Consider the following linear equations:
$ax+by+cz=0$,$bx+cy+az=0$,$cx+ay+bz=0$
Match the conditions/expressions in Column $I$ with statements in Column $II$:

Column $I$	Column $II$
$(A)$ $a+b+c \neq 0$ and $a^2+b^2+c^2=ab+bc+ca$	$(p)$ The equations represent planes meeting only at a single point.
$(B)$ $a+b+c=0$ and $a^2+b^2+c^2 \neq ab+bc+ca$	$(q)$ The equations represent the line $x=y=z$.
$(C)$ $a+b+c \neq 0$ and $a^2+b^2+c^2 \neq ab+bc+ca$	$(r)$ The equations represent identical planes.
$(D)$ $a+b+c=0$ and $a^2+b^2+c^2=ab+bc+ca$	$(s)$ The equations represent the whole of the three-dimensional space.

If matrix $A = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$,then:

If $D_1$ and $D_2$ are two $3 \times 3$ diagonal matrices,then

If $A+B=\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2\end{array}\right]$ and $AB=\left[\begin{array}{lll}1 & 2 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 1\end{array}\right]$, then $A^2+B(A+B)=$

Let $A = \begin{pmatrix} 1 & 2 & 7 \\ 4 & -2 & 8 \\ 3 & 8 & -7 \end{pmatrix}$ and $\det(A - \alpha I) = 0$, where $\alpha$ is a real number. If the largest possible value of $\alpha$ is $p$, then the circle $(x - p)^2 + (y - 2p)^2 = 320$ intersects the coordinate axes at