Answer
\begin{array}{l}
(a)=19 \\
(b)=-9 \\
(c)=3 \\
(d)=3+a^{6} \\
(c)=3-t^{8} \\
(f)=\left(a^{2}-b^{2}\right)(a-b) b^{3}+3
\end{array}
Work Step by Step
Given:
\[
x y^{2} z^{3}+3=f(x, y, z)
\]
From $13.1 .1,$ we know we have to calculate $f(2,1,2),$ substituting $x, y$ and $z$ in the function:
\[
\begin{aligned}
\because x y^{2} z^{3}+3=f(x, y, z) & \\
2 * 1^{2} * 2^{3}+3=f(2,1,2) & \\
19=f(2,1,2) &
\end{aligned}
\]
(b) Similarly substituting,
\[
(-3) * 2^{2} * 1^{3}+3=f(-3,2,1)
\]
\[
-9=f(-3,2,1)
\]
(c) Similarly substituting
\[
\begin{array}{l}
0 * 0^{2} * 0^{3}+3=f(0,0,0) \\
3=f(0,0,0)
\end{array}
\]
(d) Similarly substituting
\[
\begin{array}{l}
f(a, a, a)=a * a^{2} * a^{3}+3 \\
f(a, a, a)=a^{6}+3
\end{array}
\]
(e) Similarly substituting
\[
f\left(t, t^{2},-t\right)=t *\left(t^{2}\right)^{2} *(-t)^{3}+3
\]
\[
f\left(t, t^{2},-t\right)=3-t^{8}
\]
(f) Similarly substituting
\[
\begin{array}{l}
f(a+b, a-b, b)=(a+b)(a-b)^{2} b^{3}+3 \\
f(a+b, a-b, b)=\left(a^{2}-b^{2}\right)(a-b) b^{3}+3
\end{array}
\]
