Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46.

Let the three consecutive positive integers be x, x + 1 and x + 2.

According to the given condition,

$x^{2}+(x+1)(x+2)=46$

$\Rightarrow x^{2}+x^{2}+3 x+2=46$

$\Rightarrow 2 x^{2}+3 x-44=0$

$\Rightarrow 2 x^{2}+11 x-8 x-44=0$

$\Rightarrow x(2 x+11)-4(2 x+11)=0$

$\Rightarrow(2 x+11)(x-4)=0$

$\Rightarrow 2 x+11=0$ or $x-4=0$

$\Rightarrow x=-\frac{11}{2}$ or $x=4$

∴ x = 4           (x is a positive integer)

When x = 4,
x + 1 = 4 + 1 = 5
x + 2 = 4 + 2 = 6

Hence, the required integers are 4, 5 and 6.