= Solution
Put $P=QQ^T$. Since $Q$ has $r$ columns, the <rank bound for a matrix product> gives $\operatorname{rank}(PA)\leq r$. The <Eckart-Young-Mirsky theorem> and part (a) then give
$$
\boxed{\|PA-A\|_2\geq\sigma_{r+1}=\|Q_*Q_*^TA-A\|_2}.
$$
One can also see the lower bound directly. The restriction of $PA$ to the $(r+1)$-dimensional <linear span> of $v_1,\ldots,v_{r+1}$ has a nonzero <null space>, because its <matrix rank> is at most $r$. For a <unit vector> $z$ in that <null space>, $\|(A-PA)z\|_2=\|Az\|_2\geq\sigma_{r+1}$ by the <singular value decomposition>.
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