Question

If the point $(2, \alpha, \beta)$ lies on the plane which passes through the points $(3,4, 2)$ and $(7, 0, 6)$ and is perpendicular to the plane $2x - 5y =15$, then $2 \alpha-3\beta$ is equal to

Answer 1

Let the normal to the required plane is , $\vec{n}$ then
$\vec{n} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ 4&-4&4\\ 2&-5&0\end{vmatrix}=20\,\hat{i}+8\,\hat{j} -12\,\hat{j}$
$\therefore $ Equation of the plane
$\left(x-3\right)\times20+\left(y-4\right)\times8+\left(z-2\right)\times\left(-12\right)=0$
$5x-15+2y-8-3z+6=0$
$5x+2y-3z-17=0 \ldots\left(1\right)$
Since, equation of plane $\left(1\right)$ passes through $ \left(2, \alpha, \beta\right)$,
then $10+2\alpha-3\beta-17=0$
$\Rightarrow 2\alpha -3\beta=7$