Question #41631

Let $P_1$ and $P_2$ be two planes given by the equations $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 4$ and $\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) = 3$, respectively. A plane $\Pi$ is chosen from the family of planes passing through the line of intersection of $P_1$ and $P_2$. A fair coin is tossed. If it shows Heads, $\Pi$ is chosen to be $\Pi_1$, which is parallel to the vector $\vec{v}_1 = 2\hat{i} + \hat{j} - \hat{k}$. If it shows Tails, $\Pi$ is chosen to be $\Pi_2$, which is parallel to the vector $\vec{v}_2 = \hat{i} - 3\hat{j}$.

A point $Q$ is randomly generated by a computer. The probability that $Q$ lies on $\Pi_1$ is $0.175$, and the probability that $Q$ lies on $\Pi_2$ is $0.186$. Given that the generated point $Q$ lies on the chosen plane $\Pi$, let $p$ be the probability that the chosen plane was $\Pi_1$.

Let $\vec{n}_1$ be the normal vector to $\Pi_1$, scaled such that its $\hat{i}$ component is $1$. A vector $\vec{a}$ has magnitude $\sqrt{26}$ and satisfies $\vec{a} \cdot \vec{n}_1 = -10$. Find the value of: $361 p + |\vec{n}_1 \times \vec{a}|^2$