Step 1: Set up the pure pressures.
The pure vapour pressures of X, Y, Z are in the ratio 3:2:1. So we can call them $3k$, $2k$, and $1k$ for some number $k$. Once we find $k$, we can find each partial pressure.
Step 2: Set up the mole fractions.
The mole fractions of X, Y, Z are in the ratio 1:2:3, and they must add up to 1. Their total parts are $1+2+3 = 6$, so the mole fractions are $\frac{1}{6}$, $\frac{2}{6}$, and $\frac{3}{6}$ for X, Y, Z.
Step 3: Apply Raoult's law for the total.
For an ideal solution the total pressure is the sum of each mole fraction times each pure pressure: \[ P_{total} = \tfrac{1}{6}(3k) + \tfrac{2}{6}(2k) + \tfrac{3}{6}(1k). \]
Step 4: Simplify the total.
Adding the parts: $\tfrac{1}{6}(3k) = \tfrac{3k}{6}$, $\tfrac{2}{6}(2k) = \tfrac{4k}{6}$, and $\tfrac{3}{6}(1k) = \tfrac{3k}{6}$. The sum is $\tfrac{3k+4k+3k}{6} = \tfrac{10k}{6} = \tfrac{5k}{3}$.
Step 5: Solve for $k$.
The mixture boils at 1.5 atm, so $P_{total} = 1.5$ atm. Setting $\tfrac{5k}{3} = 1.5$ gives $k = \tfrac{1.5 \times 3}{5} = \tfrac{4.5}{5} = 0.9$ atm.
Step 6: Find the partial pressure of Y.
The partial pressure of Y is its mole fraction times its pure pressure: $p_Y = \tfrac{2}{6} \times 2k = \tfrac{2}{6} \times 2(0.9) = \tfrac{2}{6} \times 1.8 = 0.6$ atm, which is $\tfrac{3}{5}$.
\[ \boxed{p_Y = \dfrac{3}{5}\ \text{atm}} \]