Explanation:
C Let, \(\quad \mathrm{x}=\mathrm{a}-\mathrm{r}, \quad \mathrm{y}=\mathrm{a}, \quad \mathrm{z}=\mathrm{a}+\mathrm{r}\)
Now, we have
\(x+y+z=-3\)
\(\therefore \quad \mathrm{a}-\mathrm{r}+\mathrm{a}+\mathrm{a}+\mathrm{r}=-3\)
\(3 a=-3\)
\(a=-1\)
\(\text { Again, } x y z=8\)
\(\therefore \quad(a-r)(a)(a+r)=8\)
\(\mathrm{a}\left(\mathrm{a}^2-\mathrm{r}^2\right)=8\)
\(-1\left(1-r^2\right)=8\)
\(-1+\mathrm{r}^2=8\)
\(\mathrm{r}^2=9\)
\(\mathrm{r}= \pm 3\)
\(\therefore \quad \mathrm{x}, \mathrm{y}, \mathrm{z} \text { are }-4,-1,2 \text { or } 2,-1,-4\)
\(\therefore \quad \mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2=(-4)^2+(-1)^2+(2)^2\)
\(=16+1+4\)
\(=21\)