This problem tests your understanding of 3D geometry, including finding the foot of a perpendicular from a point to a plane, calculating distances, and using the area of a triangle.

1. Key Concepts and Formulas

   • Distance from a Point to a Plane: The perpendicular distance from a point $A(x_0, y_0, z_0)$ to a plane $P: ax+by+cz+d=0$ is given by $d = \frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}$. This distance also represents the length of the perpendicular segment AN.
   • Foot of the Perpendicular (N): The line segment AN is perpendicular to the plane P. Its direction ratios are the same as the normal vector to the plane, $(a,b,c)$. The coordinates of N can be found by parameterizing the line AN and finding its intersection with the plane P.
   • Area of a Right-Angled Triangle: If N is the foot of the perpendicular from A to the plane P, and B is a point on the plane P, then the line segment AN is perpendicular to any line segment NB lying in the plane. Therefore, $\triangle ABN$ is a right-angled triangle at N. Its area is given by $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BN \times AN$.
   • Distance Formula in 3D: The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

2. Step-by-Step Solution

   Step 1: Calculate the length of AN (distance from A to plane P). The point A is $(4,3,1)$ and the plane P is $x-y+2z+3=0$. The length of AN is the perpendicular distance from A to P. $AN = \frac{|(1)(4) + (-1)(3) + (2)(1) + 3|}{\sqrt{(1)^2 + (-1)^2 + (2)^2}}$ $AN = \frac{|4 - 3 + 2 + 3|}{\sqrt{1 + 1 + 4}}$ $AN = \frac{|6|}{\sqrt{6}} = \sqrt{6}$ So, the length of AN is $\sqrt{6}$ units.

   Step 2: Find the coordinates of N (foot of the perpendicular). The line AN passes through A$(4,3,1)$ and is perpendicular to the plane $x-y+2z+3=0$. The direction ratios of the normal to the plane are $(1, -1, 2)$, which are also the direction ratios of the line AN. The parametric equation of the line AN is: $\frac{x-4}{1} = \frac{y-3}{-1} = \frac{z-1}{2} = \lambda$ Any point on this line can be represented as $(4+\lambda, 3-\lambda, 1+2\lambda)$. Since N lies on the plane P, we substitute these coordinates into the plane's equation: $(4+\lambda) - (3-\lambda) + 2(1+2\lambda) + 3 = 0$ $4+\lambda - 3+\lambda + 2+4\lambda + 3 = 0$ $6\lambda + 6 = 0 \implies \lambda = -1$ Substitute $\lambda = -1$ back into the parametric coordinates to find N: $N = (4+(-1), 3-(-1), 1+2(-1)) = (3, 4, -1)$ The coordinates of the foot of the perpendicular N are $(3,4,-1)$.

   Step 3: Use the area of $\triangle ABN$ to find the length of BN. Since N is the foot of the perpendicular from A to plane P, and B lies on plane P, $\triangle ABN$ is a right-angled triangle at N. The area of $\triangle ABN$ is given as $3\sqrt{2}$. $\text{Area} = \frac{1}{2} \times AN \times BN$ $3\sqrt{2} = \frac{1}{2} \times \sqrt{6} \times BN$ $BN = \frac{2 \times 3\sqrt{2}}{\sqrt{6}} = \frac{6\sqrt{2}}{\sqrt{2}\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3}$ To simplify calculations with the distance formula, we find $BN^2$: $BN^2 = (2\sqrt{3})^2 = 12$

   Step 4: Establish a relation between $\alpha$ and $\beta$ and use the distance formula for BN. Point $B(5, \alpha, \beta)$ lies on the plane P: $x-y+2z+3=0$. So, its coordinates must satisfy the plane's equation: $5 - \alpha + 2\beta + 3 = 0$ $8 - \alpha + 2\beta = 0 \implies \alpha = 2\beta + 8 \quad \text{(i)}$ Now, use the distance formula for $BN^2$ with $B(5, \alpha, \beta)$ and $N(3, 4, -1)$: $BN^2 = (5-3)^2 + (\alpha-4)^2 + (\beta-(-1))^2$ $12 = (2)^2 + (\alpha-4)^2 + (\beta+1)^2$ $12 = 4 + (\alpha-4)^2 + (\beta+1)^2$ $8 = (\alpha-4)^2 + (\beta+1)^2 \quad \text{(ii)}$

   Step 5: Solve for $\alpha$ and $\beta$. Substitute the relation $\alpha = 2\beta+8$ from (i) into (ii): $8 = ((2\beta+8)-4)^2 + (\beta+1)^2$ $8 = (2\beta+4)^2 + (\beta+1)^2$ Expand the terms: $8 = (4\beta^2 + 16\beta + 16) + (\beta^2 + 2\beta + 1)$ Combine like terms: $8 = 5\beta^2 + 18\beta + 17$ $5\beta^2 + 18\beta + 9 = 0$ This is a quadratic equation in $\beta$. We can factor it: $5\beta^2 + 15\beta + 3\beta + 9 = 0$ $5\beta(\beta+3) + 3(\beta+3) = 0$ $(5\beta+3)(\beta+3) = 0$ This yields two possible values for $\beta$: $\beta = -3 \quad \text{or} \quad \beta = -\frac{3}{5}$ The problem states that $\alpha, \beta \in \mathbb{Z}$ (integers). Therefore, we must choose $\beta = -3$. Substitute $\beta = -3$ back into equation (i) to find $\alpha$: $\alpha = 2(-3) + 8 = -6 + 8 = 2$ So, the integer values are $\alpha=2$ and $\beta=-3$.

   Step 6: Calculate the final expression. We need to find the value of $\alpha^2 + \beta^2 + \alpha\beta$. Substitute $\alpha=2$ and $\beta=-3$: $\alpha^2 + \beta^2 + \alpha\beta = (2)^2 + (-3)^2 + (2)(-3)$ $= 4 + 9 - 6$ $= 13 - 6 = 7$

3. Common Mistakes & Tips

   • Geometric Interpretation: Always clearly visualize the geometry. N being the foot of the perpendicular means $\triangle ABN$ is right-angled at N.
   • Algebraic Errors: Be careful with expanding squares and solving quadratic equations. Double-check your arithmetic, especially when substituting values.
   • Constraints: Pay close attention to constraints like $\alpha, \beta \in \mathbb{Z}$. This often helps in uniquely determining values when multiple solutions arise from quadratic equations.
   • Distance vs. Square of Distance: When using the distance formula in equations, it's often easier to work with the square of the distance ($D^2$) to avoid square roots until the final step.

4. Summary We first calculated the length of the perpendicular AN from point A to plane P and found the coordinates of N. Recognizing that $\triangle ABN$ is a right-angled triangle at N, we used its given area to determine the length of BN. Then, using the fact that B lies on the plane, we established a relation between $\alpha$ and $\beta$. Finally, by equating the calculated $BN^2$ with the distance formula using the coordinates of B and N, we formed a quadratic equation for $\beta$. Solving it and applying the integer constraint for $\alpha$ and $\beta$, we found $\alpha=2$ and $\beta=-3$. Substituting these values into the required expression $\alpha^2+\beta^2+\alpha\beta$ yielded the final answer.

The final answer is $\boxed{7}$.