Let $a,$ $b,$ $c,$ $d$ be positive real numbers such that
\begin{align*}
(a + b)(c + d) &= 143, \\
(a + c)(b + d) &= 150, \\
(a + d)(b + c) &= 169.
\end{align*}Find the smallest possible value of $a^2 + b^2 + c^2 + d^2.$
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 $\boxed{214}.$