Let equation of #L1# be #y=x-5# , which has a slope #m# of #1#.

and equation of #color(red)(L3)# be #color(red)(y=-2x-6)#

Draw a line from #A(2,3)#, parallel to #L1#, to meet #L3# at #P#, as shown in the figure.

#AP# is the distance measured parallel to #L1# from #L3#

Let the line joining #A# and #P# be #L2#.

Given that #L2# is parallel to #L1#,

#=> L2# has the same slope #(m=1)# as that of #L1#,

Now find the equation of #L2# through #A(2,3)# with a slope of #m=1#.

#y=mx+b#

#=> 3=1xx2+b, => b=1#

#=># equation of #color(red)(L2)# in slope-intercept form is #color(red)(y=x+1)#

Set the equations of #L2 and L3# equal to each other to find the intersection point #P#

#=> x+1=-2x-6, => x=-7/3#

#=> y=-(7/3)+1=-4/3#

#=># coordinates of #P = (-7/3,-4/3)#

The distance from #A(2,3)# to #P(-7/3,-4/3)# is

#d=sqrt((-7/3-2)^2+(-4/3-3)^2)#

#=sqrt(338/9)~~6.13# units