For what value of $p$ does the system of equations $18x+36y+45=0$ and $px-54y+67=0$ have no solution?

For a system of linear equations $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ to have no solution, the condition is:
$$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$$
From the given equations:$a_1=18$, $b_1=36$, $c_1=45$$a_2=p$, $b_2=-54$, $c_2=67$ Equating the first two ratios:
$$\frac{18}{p}=\frac{36}{-54}$$
Simplify the right side:
$$\frac{18}{p}=\frac{2}{-3}$$
Cross-multiplying to solve for $p$:
$$2p=18\times (-3)$$
$$2p=-54$$
$$p=-27$$
Since $\frac{18}{-27}=-\frac{2}{3}$ and $\frac{45}{67}\ne -\frac{2}{3}$, the condition for no solution is perfectly satisfied.