Reduce the polynomial to the standard form, combining like terms

    3x^4 - 8x^3 + 5x^2 + 2x - 17 - 2x^4 + 10x^3 + 11x^2 + 18x - 14 = x^4 + 2x^3 + 16x^2 + 20x - 31.


Notice that 

    p^2 + q^2 + r^2 + s^2 = (p + q + r + s)^2 - 2*(pq + pr +ps + qr + qs + rs)   (standard decomposition for the square of a sum)


Also notice that the leading coefficient of the polynomial standard form at x^4 is 1.


Due to Vieta's theorem

    pq + pr + ps + qr + qs + rs = 16  (the sum of in-pairs products of the roots is the coefficient at x^2)

    p + q + r + s =  -2               (the sum of the roots is the coefficient at x^3 with the opposite sign).


Therefore,

    p^2 + q^2 + r^2 + s^2 = (-2)^2 - 2*16 = 4 - 32 = -28.


ANSWER.  p^2 + q^2 + r^2 + s^2 = -28.   

         (by the way, since the sum of squares is negative, it means that some roots are complex numbers).