Let P be a point in the first octant, whose image Q in the plane $x + y = 3 ($that is, the line segment PQ is perpendicular to the plane $x + y = 3$ and the mid-point of PQ lies in the plane $x + y = 3)$ lies on the z-axis. Let the distance of P from the x-axis be 5. If R is the image of P in the xy-plane, then the length of PR is _____.

$\text{Given: } P \text{ in first octant, image of } P \text{ in plane } x+y=3 \text{ is } Q \text{ on } z\text{-axis}$ $PQ \perp \text{plane } x+y=3, \quad \text{midpoint of } PQ \text{ lies in the plane}$ $d(P,\ x\text{-axis}) = 5, \qquad R = \text{image of } P \text{ in the } xy\text{-plane}$ $\text{To find: length } PR = ?$ $1.\ P=(a,b,c)$ $2.\ \text{image in } x+y=3 \Rightarrow Q,\ \ Q \text{ on } z\text{-axis} \Rightarrow a,b$ $3.\ d(P,x\text{-axis})=5 \Rightarrow c \qquad 4.\ R \Rightarrow PR$ $P=(a,b,c),\qquad a,b,c>0$ $\text{Plane: } x+y=3$ $\vec n=(1,1,0),\qquad \vec n\cdot P=a+b,\qquad |\vec n|^2=2$ $Q=P-2\,\dfrac{\vec n\cdot P-3}{|\vec n|^2}\,\vec n$ $Q=(a,b,c)-(a+b-3)(1,1,0)$ $Q=(3-b,\ 3-a,\ c)$ $Q \text{ on } z\text{-axis} \Rightarrow 3-b=0,\ \ 3-a=0$ $a=3,\qquad b=3$ $\text{mid}(PQ)=\left(\dfrac{a+3-b}{2},\dfrac{b+3-a}{2},c\right)=\left(\dfrac{3}{2},\dfrac{3}{2},c\right)$ $x+y=3\ \ \checkmark$ $d(P,x\text{-axis})=\sqrt{b^2+c^2}=\sqrt{9+c^2}=5$ $9+c^2=25\ \Rightarrow\ c^2=16$ $c=4\ \ (\text{first octant})$ $P=(3,3,4)$ $R=(3,3,-4)$ $PR=|4-(-4)|=8$ $\boxed{PR=8}$