Let two vertices of a triangle ABC be (2, 4, 6) and (0, - 2, - 5) ,...

Given,

Two vertices of a triangle $ABC$ be \left(2, 4, 6\right)$and \left(0, - 2, - 5\right)$, and its centroid be \left(2, 1, - 1\right)$</p><p>Now, let the vertex '$C$' be \left(a, b, c\right)$ and using centroid formula we get,

$2 = \dfrac{2 + 0 + a}{3} ⇒ a = 4$

$1 = \dfrac{4 - 2 + b}{3} ⇒ b = 1$

$- 1 = \dfrac{6 - 5 + c}{3} ⇒ c = - 4$

So, the third vertex will be, $C\left(4, 1, - 4\right)$

Now finding image of $C$ in $x + 2y + 4z = 11$ we get,

$\dfrac{\alpha - 4}{1} = \dfrac{\beta - 1}{2} = \dfrac{\gamma + 4}{4} = - 2\dfrac{\left(4 + 2 - 16 - 11\right)}{16 + 4 + 1}$

$⇒ \dfrac{\alpha - 4}{1} = 2 ⇒ \alpha = 6$, $\dfrac{\beta - 1}{2} = 2 ⇒ \beta = 5$ and $\dfrac{\gamma + 4}{4} = 2 ⇒ \gamma = 4$

Now find the required answer we get,

$\alpha\beta + \beta\gamma + \alpha\gamma = 30 + 20 + 24 = 74$