The planes and have equations and respectively. Verify that the point lies on the plane . Determine…

The equation of plane $\pi_1$ is $2x + 6y - 2z = 5$. Substituting the coordinates of point $P(2, \frac{1}{2}, 1)$ into the LHS: $\text{LHS} = 2(2) + 6\left(\frac{1}{2}\right) - 2(1) = 4 + 3 - 2 = 5 = \text{RHS}$. Therefore, point P lies on the plane $\pi_1$.

For planes $\pi_1: 2x + 6y - 2z = 5$ and $\pi_2: 3x + 9y + pz = -\frac{51}{2}$ to be parallel, their normal vectors must be proportional: $\frac{3}{2} = \frac{9}{6} = \frac{p}{-2}$. Since $\frac{3}{2} = \frac{9}{6}$, we have $\frac{p}{-2} = \frac{3}{2} \Rightarrow p = -3$.

A normal to $\pi_1$ has direction ratios $(2, 6, -2)$, which can be simplified to $(1, 3, -1)$. The equation of the line passing through $P(2, \frac{1}{2}, 1)$ in the direction of the normal is: $\frac{x-2}{1} = \frac{y-1/2}{3} = \frac{z-1}{-1} = \lambda$. Any point on this line has coordinates $(2 + \lambda, \frac{1}{2} + 3\lambda, 1 - \lambda)$. Since Q is the point where this line meets $\pi_2: 3x + 9y - 3z = -\frac{51}{2}$: $3(2 + \lambda) + 9\left(\frac{1}{2} + 3\lambda\right) - 3(1 - \lambda) = -\frac{51}{2}$. Dividing by 3 gives $(2 + \lambda) + 3\left(\frac{1}{2} + 3\lambda\right) - (1 - \lambda) = -\frac{17}{2}$. $2 + \lambda + \frac{3}{2} + 9\lambda - 1 + \lambda = -\frac{17}{2} \Rightarrow 11\lambda + \frac{5}{2} = -\frac{17}{2} \Rightarrow 11\lambda = -\frac{22}{2} = -11 \Rightarrow \lambda = -1$. Substituting $\lambda = -1$ into the coordinates: $x = 2 - 1 = 1$, $y = \frac{1}{2} + 3(-1) = -\frac{5}{2}$, $z = 1 - (-1) = 2$. Therefore, the coordinates of Q are $(1, -\frac{5}{2}, 2)$.