The foci of the conic $25 x^2+16 y^2-150 x=175$ are

Transform the equation to standard ellipse form:

Starting with $25x^2 + 16y^2 - 150x = 175$, complete the square in $x$:
$25(x^2 - 6x) + 16y^2 = 175$
$25(x - 3)^2 + 16y^2 = 400$

Divide by 400 to obtain standard ellipse form:
$\frac{(x-3)^2}{16} + \frac{y^2}{25} = 1$

Identify ellipse parameters:

The center is at $(3, 0)$.
Since $25 > 16$, the semi-major axis $a = 5$ and semi-minor axis $b = 4$ along the $y$ and $x$ directions respectively.

The focal distance is $c = \sqrt{a^2 - b^2} = \sqrt{25 - 16} = 3$.

Locate the foci:

With major axis vertical, the foci are located at $(h, k \pm c) = (3, \pm 3)$.

The correct option is $(3, \pm 3)$.