<s>[INST] Area of the base of the prism formed by three planes 
The three planes $P_1: kx + y+ z=2$,$P_2:x+y-z=3$,$P_3: x+2z=2$ form a triangular prism and area of the normal section (where the normal section of the triangular prism is the plane parallel to the base of the triangular prism) be $k_1$. Then the value of $2\sqrt{14}(k.k_1)$ is?

Since the three planes form a prism, that means the normal vectors have to be co-planar. So, $[\vec{n_1} \vec {n_2} \vec {n_3}] = 0$ (scalar triple product)
Where, $\vec{n_i}$ is the normal vector of the plane $P_i$
Solving that gave me the value $k=2$
The problem arises in evaluating $k_1$. Assuming that the fourth plane is the base of the prism, how do I find the area of the base?
By taking the cross product of $\vec{n_2}$ and $\vec{n_3}$, we get the vector perpendicular to the base of the prism. The magnitude of this vector may or may not give me the area of the base because it may give me a multiple of the area and not the true area.
Any help would be appreciated.
 [/INST] The three plane equations are $2x+y+z-2=0, x+y-z-3=0, x+2z-2=0$.
Consider the family of planes passing through the line of intersection of the first two planes.
$2x+y+x-2 + \lambda(x+y-z-3)=0$. The plane part of this family, and parallel to the third plane is:
$x+2z+1=0$.
The perpendicular distance between this plane and the third plane is $\frac{3}{\sqrt{5}}$.
The angle between the first and third plane , and between the second and third plane respectively is $\sin^{-1}(\frac{\sqrt{14}}{\sqrt{30}})$ and $\sin^{-1}(\frac{\sqrt{14}}{\sqrt{15}})$.
So we can easily find two side lengths of the triangle enclosed as $\frac{3\sqrt{30}}{\sqrt{75}}$ and $\frac{3\sqrt{15}}{\sqrt{70}}$.
The angle between these two sides is the angle between the first and second plane is $\sin^{-1}(\frac{\sqrt{14}}{3 \sqrt{2}})$.
So the area of the triangle is $k_1=\frac{1}{2} \times \frac{3\sqrt{15}}{\sqrt{70}} \times \frac{3\sqrt{30}}{\sqrt{70}} \times (\frac{\sqrt{14}}{3 \sqrt{2}})=\frac{9}{2\sqrt{14}}$.
So the final answer is $2\sqrt{14} \times 2 \times \frac{9}{2\sqrt{14}} = \boxed{18}$.
PS: I got the solution after placing the bounty lol :)
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