Question 12

Given that the scalar triple product $(\vec{OP} \times \vec{OQ}) \cdot \vec{OR} = 0$, the vectors are coplanar. This implies the determinant of their components is zero: $\begin{vmatrix} \frac{\alpha-1}{\alpha} & 1 & 1 \\ 1 & \frac{\beta-1}{\beta} & 1 \\ 1 & 1 & \frac{1}{2} \end{vmatrix} = 0$ Let $a = 1 - \frac{1}{\alpha}$ and $b = 1 - \frac{1}{\beta}$. The determinant becomes: $\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & 1/2 \end{vmatrix} = 0$ Expanding the determinant: $a\left(\frac{b}{2} - 1\right) - 1\left(\frac{1}{2} - 1\right) + 1(1 - b) = 0$ $\frac{ab}{2} - a + \frac{1}{2} + 1 - b = 0$ $ab - 2a - 2b + 3 = 0$ Substitute $a = 1 - \frac{1}{\alpha}$ and $b = 1 - \frac{1}{\beta}$: $\left(1 - \frac{1}{\alpha}\right)\left(1 - \frac{1}{\beta}\right) - 2\left(1 - \frac{1}{\alpha}\right) - 2\left(1 - \frac{1}{\beta}\right) + 3 = 0$ $1 - \frac{1}{\beta} - \frac{1}{\alpha} + \frac{1}{\alpha\beta} - 2 + \frac{2}{\alpha} - 2 + \frac{2}{\beta} + 3 = 0$ $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\alpha\beta} = 0$ Multiplying by $\alpha\beta$, we get $\beta + \alpha + 1 = 0$, so $\alpha + \beta = -1$. The point $(\alpha, \beta, 2)$ lies on the plane $3x + 3y - z + l = 0$, so: $3\alpha + 3\beta - 2 + l = 0$ $3(\alpha + \beta) - 2 + l = 0$ $3(-1) - 2 + l = 0 \implies -3 - 2 + l = 0 \implies l = 5.$