Let the plane x + 3y - 2z + 6 = 0 meet the co-ordinate axes at the...

Given,

The plane $x + 3y - 2z + 6 = 0$ meet the co-ordinate axes at the points $A, B, C$,

So, $A\left(- 6, 0, 0\right), B\left(0, - 2, 0\right) \text{ and } C\left(0, 0, 3\right)$

And the orthocenter of the triangle $ABC$ is \left(\alpha, \beta, \dfrac{6}{7}\right)$,</p><p>We know that, circumcentre of triangle is point of intersection of perpendicular bisectors,</p><p>So, plotting the diagram we get,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/b8dab166-7883-4d34-8fc8-2b183a0b4e7f.png" style="width: 300px; height: 264px;" /></p><p>Now from diagram we can see that$O\text{to midpoint of}BC ⊥ BC$</p><p>So, by perpendicular vector formula we get,</p><p>\left(x - 0\right) \times 0 + \left(y + 1\right) \times 2 + \left(z - \dfrac{3}{2}\right) \times 3 = 0$

$⇒ 4y + 6z - 5 = 0 . . . . . . . \left(1\right)$

Similarly, for side $AC$ we get,

\left(x + 3\right) \times \left(- 6\right) + \left(y - 0\right) \times 0 + \left(z - \dfrac{3}{2}\right)\left(- 3\right) = 0$</p><p>$⇒ 4x + 2z + 9 = 0 . . . . . . . . . \left(2\right)$</p><p>Similarly, for side$AB$we get,</p><p>$⇒ \left(x + 3\right) \times \left(- 6\right) + \left(y + 1\right) \times 2 + \left(z - 0\right) \times 0 = 0$</p><p>$⇒ 3x - y + 8 = 0 . . . . . . . . . \left(3\right)$</p><p>Now from equation \left(1\right), \left(2\right) \text{ and } \left(3\right)$ we get,

$x = \dfrac{- 9}{4} - \dfrac{z}{2}, y = \dfrac{5}{4} - \dfrac{3}{2}z, z = z$

Now using the relation between centroid, circumcentre and orthocentre, we get,

$⇒ G = \left(\dfrac{\alpha - \dfrac{9}{2} - z}{3}, \dfrac{\beta + \dfrac{5}{2} - 3z}{3}, \dfrac{\dfrac{6}{7} + 2z}{3}\right)$

$⇒ \left(- 2, - \dfrac{2}{3}, 1\right) = \left(\dfrac{\alpha - \dfrac{9}{2} - z}{3}, \dfrac{\beta + \dfrac{5}{2} - 3z}{3}, \dfrac{\dfrac{6}{7} + 2z}{3}\right)$

Now on comparing both side we get,

$⇒ z = \dfrac{15}{14}, \alpha = - \dfrac{3}{7}, \beta = - \dfrac{9}{7}$

$⇒ 98{\left(\alpha + \beta\right)}^{2} = 98 \times \dfrac{144}{49} = 288$