There exist three numbers in the ratio of 3 : 2 : 7 such that the sum of their squares is equal to 558. The three numbers will be respectively:

The three numbers are in the ratio 3 : 2 : 7.

Let the common multiplier be $x$.

The three numbers are:

First number $= 3x$

Second number $= 2x$

Third number $= 7x$

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The sum of their squares equals 558:

$(3x)^2 + (2x)^2 + (7x)^2 = 558$

$9x^2 + 4x^2 + 49x^2 = 558$

$62x^2 = 558$

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$x^2 = \dfrac{558}{62}$

$x^2 = 9$

$x = 3$

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The three numbers are:

First number $= 3x = 3 \times 3 = 9$

Second number $= 2x = 2 \times 3 = 6$

Third number $= 7x = 7 \times 3 = 21$

Therefore, the three numbers are 9, 6, and 21 respectively.