Let the foot of perpendicular from the point A(4, 3, 1) on the...

Given,

The foot of perpendicular from the point $A\left(4, 3, 1\right)$ on the plane $P: x - y + 2z + 3 = 0$ be $N$

So, by foot of perpendicular formula we get,

$\dfrac{{x}_{1} - 4}{1} = \dfrac{{y}_{1} - 3}{- 1} = \dfrac{{z}_{1} - 1}{2} = - \dfrac{4 - 3 + 2 + 3}{{1}^{2} + {1}^{2} + {2}^{2}}$

$⇒ N ≡ \left({x}_{1}, {y}_{1}, {z}_{1}\right) = ( 3, 4, - 1 )$

And $B\left(5, \alpha, \beta\right), \alpha, \beta \in \mathbb{Z}$ is a point on plane $P$ such that the area of the triangle $ABN$ is $3\sqrt{2}$,

So putting the value of point $B\left(5, \alpha, \beta\right)$ in plane we get,

$5 - \alpha + 2\beta + 3 = 0$

$⇒ \alpha - 2\beta = 8$

And given area of triangle $ABN$ is $3\sqrt{2}$, we get

$\dfrac{1}{2}AN \times BN = 3\sqrt{2}$

$⇒ AN \times BN = 6\sqrt{2}$

$⇒ \dfrac{6}{\sqrt{6}} \times BN = 6\sqrt{2}$

$⇒ BN = 2\sqrt{3}$

$⇒ 4 + {\left(\alpha - 4\right)}^{2} + {\left(\beta + 1\right)}^{2} = 12$

$⇒ {\left(\alpha - 4\right)}^{2} + {\left(\beta + 1\right)}^{2} = 8$

$⇒ 5{\beta}^{2} + 18\beta + 9 = 0$ \left\{\text{as}\alpha - 2\beta = 8\right\}$</p><p>$⇒ \beta = - 3\left\{\text{as}\beta \ne - \dfrac{3}{5}\right\}$</p><p>So,$\beta = - 3, \alpha = 2$</p><p>Hence,${\alpha}^{2} + {\beta}^{2} + \alpha\beta = {3}^{2} + {2}^{2} - 6 = 7