Question:

Step 1: Parametrize the line containing edge $CD$. The line is given by $\frac{x-1}{1} = \frac{y-1}{2} = \frac{z-3}{3}$. Let this common ratio be $t$. Then any point on the line can be expressed as $(1+t, 1+2t, 3+3t)$. Let $C = (1+t_1, 1+2t_1, 3+3t_1)$ and $D = (1+t_2, 1+2t_2, 3+3t_2)$ for some parameters $t_1, t_2 \in \mathbb{R}$. Step 2: Determine the coordinates of the centroid $G$ of the tetrahedron $ABCD$. The vertices are $A = (2, 0, 0)$, $B = (0, 4, 0)$, $C = (1+t_1, 1+2t_1, 3+3t_1)$, and $D = (1+t_2, 1+2t_2, 3+3t_2)$. The centroid $G = (G_x, G_y, G_z)$ is given by the average of the coordinates of its vertices: $$G = \frac{A+B+C+D}{4}$$ Step 3: Compute each coordinate of $G$. $$G_x = \frac{2+0+(1+t_1)+(1+t_2)}{4} = \frac{4+t_1+t_2}{4} = 1 + \frac{t_1+t_2}{4}$$ $$G_y = \frac{0+4+(1+2t_1)+(1+2t_2)}{4} = \frac{6+2(t_1+t_2)}{4} = \frac{3}{2} + \frac{t_1+t_2}{2}$$ $$G_z = \frac{0+0+(3+3t_1)+(3+3t_2)}{4} = \frac{6+3(t_1+t_2)}{4} = \frac{3}{2} + \frac{3(t_1+t_2)}{4}$$ Step 4: Express the locus of $G$ using a single parameter. Let $s = \frac{t_1+t_2}{4}$. Substituting $s$ into the expressions for $G_x, G_y, G_z$: $$G_x = 1+s$$ $$G_y = \frac{3}{2}+2s$$ $$G_z = \frac{3}{2}+3s$$ From these equations, we can express $s$ in terms of each coordinate: $$s = G_x - 1$$ $$s = \frac{G_y - \frac{3}{2}}{2}$$ $$s = \frac{G_z - \frac{3}{2}}{3}$$ Thus, the locus of the centroid $G$ is the line given by: $$\frac{x-1}{1} = \frac{y-\frac{3}{2}}{2} = \frac{z-\frac{3}{2}}{3}$$ Step 5: Match the derived locus with the given form to find $y_1, z_1, a, b$. The given form of the locus is $\frac{x - \frac{3}{2}}{1} = \frac{y - y_1}{a} = \frac{z - z_1}{b}$. The derived locus is $\frac{x-1}{1} = \frac{y-\frac{3}{2}}{2} = \frac{z-\frac{3}{2}}{3}$. Both equations represent the same line. The direction vector of this line is $(1, 2, 3)$. Comparing the direction vectors, we have $a=2$ and $b=3$. To find $y_1$ and $z_1$, we need to find the point on the derived locus where the $x$-coordinate is $\frac{3}{2}$. From $G_x = 1+s$, if $G_x = \frac{3}{2}$, then $1+s = \frac{3}{2}$, which implies $s = \frac{1}{2}$. Substitute $s = \frac{1}{2}$ into the expressions for $G_y$ and $G_z$: $$G_y = \frac{3}{2} + 2\left(\frac{1}{2}\right) = \frac{3}{2} + 1 = \frac{5}{2}$$ $$G_z = \frac{3}{2} + 3\left(\frac{1}{2}\right) = \frac{3}{2} + \frac{3}{2} = 3$$ So, the point $\left(\frac{3}{2}, \frac{5}{2}, 3\right)$ lies on the locus. Comparing this point with $\left(\frac{3}{2}, y_1, z_1\right)$ from the given form, we find $y_1 = \frac{5}{2}$ and $z_1 = 3$. The values are $a=2$, $b=3$, $y_1=\frac{5}{2}$, and $z_1=3$.