Find distance of the point (-2,3,-4) from the line (x+2)/3=(2y+3)/4=(3z+4)/5, measured parallel to the plane 4x+12y-3z+1=0

Equation of given line is (x+2)/3=(2y+3)/4=(3z+4)/5=λ (say)

Therefore Any point on this line is of the form, 

[3λ-2,(4λ-3)/2, (5λ-4)/3]

Now, D.R.S of the line from the point (-2,3,-4) is 3λ,(4λ-9)/2,(5λ+8)/3

Also the Equation of Plane is given as 4x+12y-3z+1=0

Therefore D.R.S of Normal= 4,12,-3

Now the normal of plane is perpendicular to the drs of above line . Hence,

4.3λ+12[(4λ-9)/2]-3[(5λ+8)/3]=0

=> 12λ+24λ-54-5λ-8=0

=> 31λ=62

=> λ=2

Hence the required Coordinate is (4,5/2,2)

Hence Distance Between Coordinates (4,5/2,2) and (-2,3,-4) is 17/2 units( By Distance Formula)

Hope it helps!!