68 Let $a>0, A=\left\{(x, y) \mid(x-3)^{2}+(y+4)^{2} \leqslant 1\right\} \quad B=\{(x, y)|| x-3 |$ $+2|y+4| \leqslant a\}$, then the necessary and sufficient condition for $A \subseteq B$ is
A. $a \geqslant 2$
B. $a \geqslant 3$
C. $a \geqslant \sqrt{3}$
D. $a \geqslant \sqrt{5}$
$68 \mathrm{D}$. Let $(x, y) \in A$, then
$$
\left\{\begin{array}{l}
x-3=t \cos \theta, \\
y+4=t \sin \theta
\end{array} \quad(0 \leqslant t \leqslant 1)\right.
$$

Substitute into
$$
|x-3|+2|y+4| \leqslant a,
$$

we get
$$
|t \cos \theta|+2|t \cdot \sin \theta| \leqslant a .
$$

Take $t=1, \theta=\arcsin \frac{2}{\sqrt{5}}$, then
$$
|t \cos \theta|+2|t \sin \theta|=\sqrt{5} \cos \left(\theta-\arcsin \frac{2}{\sqrt{5}}\right)=\sqrt{5}
$$

So $a \geqslant \sqrt{5}$. And when $a \geqslant \sqrt{5}$,
$$
|t \cos \theta|+2|t \sin \theta| \leqslant|\cos \theta|+2|\sin \theta| \leqslant \sqrt{5} \leqslant a .
$$