Inspired by Nihar Mahajan

$H$ is the Orthocenter of ABC

Well , This is a pure geometry proof. I think the problem poser is inspired by this problem . The triangle with minimum perimeter is indeed the orthic triangle of $\Delta ABC$ (this can be proved). Please revise the orthic configuration before seeing this proof , or you will not understand some steps.I t almost took 2 hours for me to solve and present the solution. Do upvote if you like.

First by Heron's formula we can compute that : $A(\Delta ABC) = 48$

$A(\Delta ABC) =\dfrac{1}{2}(AB)(PC) =\dfrac{1}{2}(12)(PC) = 48\\\Rightarrow PC=8 \quad AC=8\sqrt{2} \quad m\angle CPA=90\\ \Rightarrow \Delta CPA \quad\text{is a 45-45-90 triangle} \Rightarrow AP=8$

By angle chase we also see that $\Delta CPA$ is also a 45-45-90 triangle.

$\Rightarrow AR = 6\sqrt{2} \quad,\quad PB = AB-AP=12-8 \Rightarrow PB =4$ Also , $RC=AC-AR=8\sqrt{2}-6\sqrt{2}=2\sqrt{2}$

By Pythagoras Theorem to $\Delta ABQ$ , we get $BQ=\dfrac{12\sqrt{5}}{5} \\ QC=BC-BQ=4\sqrt{5}-\dfrac{12\sqrt{5}}{5} \Rightarrow QC =\dfrac{8\sqrt{5}}{5}$

Now since we have computed all the parts of sides of $\Delta ABC$ ,we can use ratios and computations to get $PQ,QR,PR$ .

$\Delta APR \sim \Delta ACB \Rightarrow \dfrac{AP}{AC}= \dfrac{PR}{CB}\Rightarrow PR = 2\sqrt{10} \\ \Delta BPQ \sim \Delta BCA \Rightarrow \dfrac{BP}{BC}= \dfrac{PQ}{CA} \Rightarrow PQ = \dfrac{8\sqrt{10}}{5}\\ \Delta CRQ \sim \Delta CBA \Rightarrow \dfrac{CR}{BC}= \dfrac{QR}{BA}\Rightarrow QR = \dfrac{6\sqrt{10}}{5}$

For $\Delta PQR$ , semiperimeter $s=\dfrac{12\sqrt{10}}{5}$

Also by heron's formula , we find $A(\Delta PQR) = \dfrac{48}{5}$

Hence , inradius $r=\dfrac{A(\Delta PQR)}{s} =\dfrac{\dfrac{48}{5}}{\dfrac{12\sqrt{10}}{5}} = \dfrac{2\sqrt{10}}{5}$

$\Rightarrow\huge\boxed{\color{#3D99F6}{r= \dfrac{2\sqrt{10}}{5} \approx 1.265}}$

I am posting its solution using Coordinate Geometry and also looking forward for more different type of solution for this question.

Now coming to the solution,

$cosA = \dfrac{b^2+c^2-a^2}{2bc}= \dfrac{128+144-80}{2\times 8\sqrt{2} \times 12} = \dfrac{1}{\sqrt{2}} \\ \implies A = 45^{\circ}$

Now let $A$ be origin. Then $B\equiv (12,0) ~\& ~C\equiv (8,8)$

$\therefore$ Line $AC\equiv ~ y=x$ & line $BC\equiv ~ 2x+y=24$

Let point $P\equiv ~ (h,0)$ . We know that minimum perimeter of $\Delta PQR$ would be the distance between the images of point $P$ about line $AC~\&~BC$ .

Images of point $P$ about line $AC~\&~BC$ are $D\equiv (0,h)~ \&~ E\equiv~ \left( \dfrac { 96-3h }{ 5 } ,\dfrac { 48-4h }{ 5 } \right)$ respectively.

So length of $DE = l = \sqrt { \dfrac { { (96-3h) }^{ 2 } }{ 25 } +\dfrac { { (48-9h) }^{ 2 } }{ 25 } } \\ \implies ~ l^2 = \dfrac { { (96-3h) }^{ 2 } }{ 25 } +\dfrac { { (48-9h) }^{ 2 } }{ 25 } \\ \implies ~ 2l\dfrac{dl}{dh} = \dfrac { 2(96-3h)\times (-3) }{ 25 } +\dfrac { 2(48-9h)\times (-9) }{ 25 }$

For $l$ to be minimum $\dfrac{dl}{dh} = 0 \\ \implies~ -6(96-3h) = 18(48-9h) \\ \therefore h = 8$

$\therefore ~ P\equiv (8,0),~ D\equiv (0,8)~\& ~E\equiv (14.4,3.2)$

So line $DE\equiv ~ 3y+x = 24$ and points $Q ~\& ~ R$ will be point of intersection of line $DE$ with line $BC~\& ~AC$ respectively.

$\therefore ~ Q\equiv(9.6,4.8)~\& ~ R\equiv(6,6)$

Now, area of $\Delta PQR =\dfrac{1}{2} \times \begin{vmatrix} { x }_{ 1 } & { y }_{ 1 } & 1 \\ { x }_{ 2 } & { y }_{ 2 } & 1 \\ { x }_{ 3 } & y_{ 3 } & 1 \end{vmatrix} = 9.6$

And semiperimeter( $s$ ) of $\Delta PQR = \dfrac{l}{2} = \dfrac{15.18}{2} = 7.59$

$\therefore$ Inradius of $\Delta PQR = \dfrac{\Delta}{s} = \dfrac{ 9.6}{7.59} \approx ~ 1.265$

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From the first link we get,
$R=\dfrac{abc}{4*area}\\ \ \ \ =\dfrac{a^2+b^2+c^2}{8(1+CosA*C0sB*CosC)}\\ Orthic\ Pedal\ \Delta\ has \ minimum\ perimeter. \therefore\ From\ second.\\ Inradius\ of\ orthic \ \Delta \ =2R*|CosA*CosB*CosC| \\ \therefore \ R=\dfrac{4\sqrt5*8\sqrt2*12}{\sqrt{(4\sqrt5+8\sqrt2+12)(4\sqrt5+8\sqrt2-12) (4\sqrt5-8\sqrt2+12) (-4\sqrt5+8\sqrt2+12) }}\\ 1+|CosA*CosB*CosC|=\dfrac{(4\sqrt5)^2+(8\sqrt2)^2+12^2}{8*R^2}\\ Inradius\ of\ orthic \ \Delta \ =2R\{\dfrac{(4\sqrt5)^2+(8\sqrt2)^2+12^2}{8*R^2} - 1\}=\Large \ \ \ \color{#D61F06}{1.2649}.$