3. If real numbers $a, b, c, d$ satisfy $5 a+6 b-7 c+4 d=1$, then the minimum value of $3 a^{2}+2 b^{2}+5 c^{2}+d^{2}$ is

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3. $\frac{15}{782}$ Detailed explanation: From the given conditions and by applying the Cauchy-Schwarz inequality,

we get $1=(5 a+6 b-7 c+4 d)^{2}=\left(\frac{5}{\sqrt{3}} \sqrt{3} a+\frac{6}{\sqrt{2}} \sqrt{2} b+\frac{-7}{\sqrt{5}} \sqrt{5} c+4 d\right)^{2}$ $\leqslant\left(\frac{25}{3}+\frac{36}{2}+\frac{49}{5}+16\right)\left(3 a^{2}+2 b^{2}+5 c^{2}+d^{2}\right)=\frac{782}{15}\left(3 a^{2}+2 b^{2}+5 c^{2}+d^{2}\right)$. Clearly, equality can be achieved, so the minimum value sought is $\frac{15}{782}$.