Area of a triangle inside a prism

Question

The question goes as follows.

Consider $3$ planes $$1)2x+y+z=3$$ $$2)x-y+2z=4$$ $$3)x+y=2$$ such that they don't intersect in a single line and form a prism.Another plane $4)$ is made through some point $P$ on line of intersection of $2),3)i.e.L_1L_2$ such that it is made perpendicular to line $L_1L_2$ intersecting line $L_3L_4$ in $Q$ and line $L_5L_6$ in $R$ such that $\triangle PQR$ is made.

Here is a figure.

The question is to find which of the following option(s) are correct if $L$ denotes area of triangle PQR

$a)\lfloor \frac 1L \rfloor \lt 11$

$b)\lfloor \frac 1L +1 \rfloor \geq 12$

$c)(\lfloor L+1 \rfloor)^{-1} \lt 3$

$d) \lfloor \frac 2L -1 \rfloor \in [18, 20]$

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I tried by finding out the line of intersection of the planes $2$ and $3$.And since the plane of the triangle is perpendicular to line $ L_1L_2 $ so the $ DR$ of plane is given by the DR of the line. Using this i could find out the DR of ths plane of triangle and by taking $x=0$ in the line of intersection of planes i could find out the equation of the plane of the triangle. But i could not find out the range of values that $ L$ can take. Any ideas? Thanks.

Answer 1

$$P=\left(\frac53,\frac13,\frac43\right)$$ $$Q=\left(1,-\frac{1}3,\frac43\right)$$ $$R=\left(1,1,0\right)$$ $$L=\frac{4\sqrt{3}}9$$

Answer 2

Take the intersection of all planes with x, y plane, i.e z=0, Now notice that the third equation is x+y=2 that means the third plane is parallel to z axis. We take value of L in (x, y) plane for compare with options .we have three lines in x, y plane:

2x+y=3

x-y=4

x+y=2

these lines intersect at A(7/3, -5/3), B(3, -1), C(1, 1). We find:

AB=a=(2^1.5)/3,

BC=b=2^1.5,

AC =c= (5^0.5)4/3

Suppose

c = AB + BC + AC = (2/3)2^0.5 + 2(2^0.5) +4/3(5^0.5)

and p=c/2

Using formula L = (p(P-a)(p-b)(p-c))^0.5 we find L = 4/3. Therefore option

3 i.e. (L+1)^-1 < 3 is correct.