jmc

Expanding the given equations, we get $$\begin{array}{rl}ac+ad+bc+bd& =143,\\ ab+ad+bc+cd& =150,\\ ab+ac+bd+cd& =169.\end{array}$$Adding the first two equations and subtracting the third equation, we get $2ad+2bc=124,$ so $ad+bc=62.$ Then $ac+bd=143-62=81,$ and $ab+cd=150-62=88.$

Now, $$\begin{array}{rl}(a+b+c+d{)}^2& =a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)\\ & =a^2+b^2+c^2+d^2+2(62+81+88)\\ & =a^2+b^2+c^2+d^2+462.\end{array}$$Thus, minimizing $a^2+b^2+c^2+d^2$ is equivalent to minimizing $a+b+c+d.$

By AM-GM, $$a+b+c+d\ge 2\sqrt{(a+d)(b+c)}=26,$$so $a^2+b^2+c^2+d^2\ge 26^2-462=214.$

To prove that 214 is the minimum, we must find actual values of $a,$ $b,$ $c,$ and $d$ such that $a^2+b^2+c^2+d^2=214.$ From the equality case for AM-GM, $a+d=b+c=13.$

Remember that $a+b+c+d=26.$ If $a+b=13+x,$ then $c+d=13-x,$ so $$169-x^2=143,$$and $x^2=26.$

If $a+c=13+y,$ then $b+d=13+y$, so $$169-y^2=150,$$and $y^2=19$.

If we take $x=\sqrt{26}$ and $y=\sqrt{19},$ then $$\begin{array}{rl}a+d& =13,\\ b+c& =13,\\ a+b& =13+\sqrt{26},\\ a+c& =13+\sqrt{19}.\end{array}$$Solving, we find $$\begin{array}{rl}a& =\frac{1}{2}(13+\sqrt{19}+\sqrt{26}),\\ b& =\frac{1}{2}(13-\sqrt{19}+\sqrt{26}),\\ c& =\frac{1}{2}(13+\sqrt{19}-\sqrt{26}),\\ d& =\frac{1}{2}(13-\sqrt{19}-\sqrt{26}).\end{array}$$We can then conclude that the minimum value of $a^2+b^2+c^2+d^2$ is $\boxed{214}.$