A.  $\left(S\right): {\left(x+2\right)}^2+{\left(y+4\right)}^2+{\left(z+3\right)}^2=1$

B. $\left(S\right): {\left(x-2\right)}^2+{\left(y-4\right)}^2+{\left(z-3\right)}^2=6$

C. $\left(S\right): {\left(x-2\right)}^2+{\left(y-4\right)}^2+{\left(z-3\right)}^2=\frac{2}{7}$

D. $\left(S\right): {\left(x-2\right)}^2+{\left(y+4\right)}^2+{\left(z+4\right)}^2=8$

A.  ${\left(x+3\right)}^2+{\left(y+1\right)}^2+{\left(z-1\right)}^2=\frac{4}{9}$

B.  ${\left(x+1\right)}^2+{\left(y+1\right)}^2+{\left(z+1\right)}^2=\frac{4}{9}$

C.  ${\left(x-3\right)}^2+{\left(y+1\right)}^2+{\left(z-1\right)}^2=\frac{4}{9}$

D.  ${\left(x-1\right)}^2+{\left(y+1\right)}^2+{\left(z-1\right)}^2=\frac{4}{9}$

A.  $\sqrt{3}+\sqrt{11}$

B.  $12\sqrt{3}$

C.  $3\sqrt{3}$

D.  $10\sqrt{3}$

A.  ${\left(x+3\right)}^2+{\left(y+1\right)}^2+{\left(z-3\right)}^2=\frac{4}{9}$

B.  ${\left(x-3\right)}^2+{\left(y+1\right)}^2+{\left(z-3\right)}^2=\frac{4}{9}$

C.  ${\left(x+3\right)}^2+{\left(y+1\right)}^2+{\left(z+3\right)}^2=\frac{4}{9}$

D.  ${\left(x-3\right)}^2+{\left(y-1\right)}^2+{\left(z+3\right)}^2=\frac{4}{9}$

A.  $\frac{x+3}{8}=\frac{y+1}{3}=\frac{z-2}{4}$

B. $\frac{x-3}{8}=\frac{y+1}{3}=\frac{z-2}{4}$

C. $\frac{x+3}{-8}=\frac{y+1}{3}=\frac{z+2}{4}$

D. $\frac{x+3}{-8}=\frac{y+1}{3}=\frac{z-2}{4}$

A.  $\frac{x+3}{8}=\frac{y+1}{3}=\frac{z-2}{4}$

B.  $\frac{x-3}{8}=\frac{y+1}{3}=\frac{z-2}{4}$

C.  $\frac{x+3}{-8}=\frac{y+1}{3}=\frac{z+2}{4}$

D.  $\frac{x+3}{-8}=\frac{y+1}{3}=\frac{z-2}{4}$