I did the following: Let $W=X+Y-3Z,$ then we have that
$$W\sim N(1+1-3,1+1+9)=N(-1,11).$$
so
$$P(X+Y\ge3Z)=P(W\ge 0)=1-P(W\le 0)=1-F_W(0)=1-\Phi\left(\frac{1}{\sqrt{11}}\right)\approx \\ \approx 1-0.618 = 0.382.$$
However, I'm in the chapter treating convolution. How can I solve this problem using this approach instead?
You may not want to do it by convolution as you would see after I lay out the solution.
Let us define a random variable $W = X+Y$ where $X$ and $Y$ are $N(1,1)$.
$$f_W(w) = \int_{-\infty}^{\infty} f_Y(y)f_X(w-y)dy$$
$$f_W(w) = \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{(y-1)^2}{2}}.\frac{1}{\sqrt{2\pi}}e^{-\frac{(w-y-1)^2}{2}}dy$$
$$f_W(w) = \int_{-\infty}^{\infty}\frac{1}{2\pi}e^{-\frac{(w^2-2w+2)}{2}}.e^{-\frac{(2y^2-2wy)}{2}}dy$$
From Wolfram Alpha. $$\int_{-\infty}^{\infty}e^{-\frac{(2y^2-2wy)}{2}}dy = \sqrt{\pi}{e^{\frac{w^2}{4}}}$$
$$f_W(w) =\frac{1}{2\sqrt{\pi}}e^{-\frac{(w^2-2w+2)}{2}}.e^{\frac{w^2}{4}}$$
$$f_W(w) = \frac{1}{2\sqrt{\pi}}e^{-\frac{(w-2)^2}{2{\sqrt{2}}^2}}$$
which is what $N(2,2)$
Now Let us define $M = W-3Z = W+U$ where $W \text{ is } N(2,2)$ and $(-3Z = U) \text{ is } N(-3,9)$
$$f_M(m) = \int_{-\infty}^{\infty} f_W(w)f_U(m-w)dw$$
$$f_M(m) = \int_{-\infty}^{\infty}\frac{1}{2\sqrt{\pi}}e^{-\frac{(w-2)^2}{4}}.\frac{1}{3\sqrt{2\pi}}e^{-\frac{(m-w+3)^2}{2.(9)}}dw$$
$$f_M(m) = \int_{-\infty}^{\infty}\frac{1}{6\sqrt{2}\pi}e^{-\frac{(2m^2+12m+54)}{36}}.e^{-\frac{(11w^2-4wm-48w)}{36}}dw$$
From Wolfram Alpha. $$\int_{-\infty}^{\infty}e^{-\frac{(11w^2-4wm-48w)}{2}}dw = \frac{6\sqrt{\pi}}{\sqrt{11}}e^{\frac{(m+12)^2}{99}}$$
$$f_M(m) = \frac{1}{\sqrt{22\pi}}e^{-\frac{(2m^2+12m+54)}{36}}.{e^{\frac{(m+12)^2}{99}}}$$
$$f_M(m) = \frac{1}{\sqrt{22\pi}}e^{-\frac{(m+1)^2}{22}}$$
which is what $N(-1,11)$.
Now tell me if you would want to do it by convolution. You can't integrate some nasty integrals.