= Solution
Assume the three standard dice are fair and <independent>. For one die $X$, a <discrete uniform distribution> on $1,\ldots,6$ gives
$$
\mathbb EX=\frac{1+2+3+4+5+6}{6}=\frac72,\qquad\mathbb EX^2=\frac{1+4+9+16+25+36}{6}=\frac{91}6.
$$
The <variance> is $\operatorname{Var}X=\mathbb EX^2-(\mathbb EX)^2=35/12$. For the sum $Q=X_1+X_2+X_3$, <linearity of expectation> and addition of <variances> for <independent random variables> give
$$
\boxed{\mathbb EQ=\frac{21}2,\qquad\operatorname{Var}Q=3\frac{35}{12}=\frac{35}4.}
$$
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