Let the mean and variance of four numbers 3, 7, x and y ( x > y )...

Mean $= 5,$ Variance $= 10$
$\dfrac{3 + 7 + x + y}{4} = 5, 10 = \dfrac{∑ {x}_{i}^{2}}{4} - {\left(∑ x\right)}^{2}$

$10 + x + y = 20, 10 = \dfrac{∑ {x}_{i}^{2}}{4} - 25$

$x + y = 10$

$y = 10 - x, 40 = ∑ {x}_{i}^{2} - 100$

$140 = ∑ {x}_{i}^{2}$

$140 = 9 + 49 + {x}^{2} + {y}^{2}$
$140 = 58 + {x}^{2} + {y}^{2}$
${x}^{2} + {y}^{2} = 82$
${\left(x + y\right)}^{2} = {x}^{2} + {y}^{2} + 2xy$
$100 - 82 = 2xy$
$18 = 2xy$
$xy = 9$
$x\left(10 - x\right) = 9$
$⇒ {x}^{2} - 10x + 9 = 0$
$= x = 1, 9$
$y = 9, 1$

As $x > y$

So, $x = 9, y = 1$

$\overset{¯}{x} = \dfrac{3 + 2x + 7 + 2y + x + y + x - y}{4}$

$⇒ \dfrac{10 + 4x + 2y}{4}$

$⇒ \dfrac{5 + 2x + y}{2}$

$⇒ \dfrac{5 + 18 + 1}{2} = \dfrac{24}{2} = 12$