Expanding the given equations, we get \begin{align*} ac + ad + bc + bd &= 143, \\ ab + ad + bc + cd &= 150, \\ ab + ac + bd + cd &= 169.\end{align*}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{align*} (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{align*}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{align*} a + d &= 13, \\ b + c &= 13, \\ a + b &= 13 + \sqrt{26}, \\ a + c &= 13 + \sqrt{19}.\end{align*}Solving, we find \begin{align*} 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{align*}We can then conclude that the minimum value of $ a^2 + b^2 + c^2 + d^2 $ is $ \boxed{214} $.