[THOUGHT] We need to compute the volume of a solid with base in xy-plane: region bounded by y = sqrt(x + x ln x), x-axis, x=1, x=2. Cross-sections perpendicular to x-axis are equilateral triangles.

For a solid with cross-sections perpendicular to x-axis, area A(x) of cross-section at x, volume = ∫_{x=a}^{b} A(x) dx.

Here base region: from x=1 to x=2, the top boundary is y = sqrt(x + x ln x) = sqrt(x(1+ln x)). The bottom boundary is y=0 (x-axis). So the width of the base of the triangle is the vertical segment from y=0 to y=sqrt(x + x ln x). But cross-sections are equilateral triangles. If the cross-section is perpendicular to x-axis, the triangle lies in a plane parallel to yz-plane. The base of the triangle is the length of the segment in the base region at that x: that length is y_top - y_bottom = sqrt(x + x ln x) - 0 = sqrt(x + x ln x). So the side length of the equilateral triangle is s = sqrt(x + x ln x).

Area of equilateral triangle with side s: A = (sqrt(3)/4) s^2.

Thus A(x) = (sqrt(3)/4) * (sqrt(x + x ln x))^2 = (sqrt(3)/4) * (x + x ln x) = (sqrt(3)/4) * x (1 + ln x).

Volume V = ∫{x=1}^{2} A(x) dx = (sqrt(3)/4) ∫{1}^{2} x (1 + ln x) dx.

Compute I = ∫{1}^{2} x (1 + ln x) dx = ∫{1}^{2} x dx + ∫_{1}^{2} x ln x dx.

Compute each integral.

First ∫ x dx from 1 to 2 = [x^2/2]_{1}^{2} = (4/2 - 1/2) = (2 - 0.5) = 1.5 = 3/2.

Second ∫ x ln x dx from 1 to 2. Use integration by parts: let u = ln x, dv = x dx => du = (1/x) dx, v = x^2/2. Then ∫ u dv = uv - ∫ v du = (ln x)(x^2/2) - ∫ (x^2/2)(1/x) dx = (x^2 ln x)/2 - (1/2) ∫ x dx = (x^2 ln x)/2 - (1/2)*(x^2/2) = (x^2 ln x)/2 - x^2/4.

Evaluate from 1 to 2: at x=2: (4 ln 2)/2 - 4/4 = 2 ln 2 - 1. At x=1: (1*ln1)/2 - 1/4 = 0 - 1/4 = -1/4. So difference = (2 ln 2 - 1) - (-1/4) = 2 ln 2 - 1 + 1/4 = 2 ln 2 - 3/4.

Thus I = 3/2 + (2 ln 2 - 3/4) = 3/2 - 3/4 + 2 ln 2 = (6/4 - 3/4) + 2 ln 2 = 3/4 + 2 ln 2.

So V = (sqrt(3)/4) * I = (sqrt(3)/4) * (3/4 + 2 ln 2) = sqrt(3) * (3/4 + 2 ln 2) / 4 = sqrt(3) * (3 + 8 ln 2) / 16.

Because (3/4 + 2 ln 2) = (3 + 8 ln 2)/4. Multiply by sqrt(3)/4 gives sqrt(3)*(3+8 ln 2)/16.

Thus V = sqrt(3)*(3 + 8 ln 2)/16. So p=3, q=8, r=16. Then p+q+r = 3+8+16 = 27.

But let's verify with Python integration to be safe.

[/THOUGHT] [PYTHON] import sympy as sp

Define variable

x = sp.symbols('x')

Define integrand: x*(1 + ln(x))

integrand = x * (1 + sp.log(x))

Compute definite integral from 1 to 2

I = sp.integrate(integrand, (x, 1, 2)) print("I =", I) print("I simplified =", sp.simplify(I))

Compute volume V = sqrt(3)/4 * I

V = sp.sqrt(3)/4 * I print("V =", V) print("V simplified =", sp.simplify(V))

Express as sqrt(3)(p + qln(2))/r

We can extract coefficients

p = 3 q = 8 r = 16 print("p+q+r =", p+q+r) [/PYTHON]

[THOUGHT] The Python output confirms the volume is $\frac{\sqrt{3}(3 + 8 \ln 2)}{16}$, so $p = 3$, $q = 8$, $r = 16$. Their sum is $3 + 8 + 16 = 27$. [/THOUGHT] $\boxed{27}$