Consider the equation of line AB is $\frac{x}{2}=\frac{y}{-3}=\frac{z}{6}$ . Through a point P(1, 2, 5)line PN is drawn perpendicular to AB and line PQ is drawn parallel to the plane  $3x+4y+5z=0$ to meet AB is Q. Then, 

(a, b, c) equation of line AB is  $\frac{x}{2}=\frac{y}{–3}=\frac{z}{6}$
Its DR’s are < 2, – 3, 6 >
Let the coordinates be < 2r, – 3r, 6r >
DR’s of PN are < 2r – 1, 3r – 2, 6r – 5 >
 It is perpendicular to AB
  2 (2r – 1)– 3(–3r – 2) + 6(6r – 5) = 0
 4r – 2 + 9r + 6 + 36r – 30 = 0
 49r = 26 i.e. $r=\frac{26}{49}$ 
(a)  Coordinates of N are  $(\frac{52}{49},–\frac{78}{49},\frac{156}{49})$
(b) Let the coordinates of Q be (2r, – 3r, 6r), then DR’s of  PQ are < 2r – 1, – 3r – 2, 6r – 5 >. Since, PQ is parallel to the plane. 
 $$\begin{gathered} \therefore 3(2r–1)+4(–3r–2)+5(6r–5)=0 \\ 6r–3–12r–8+30r–25=0 \\ 24r=36,r=\frac{3}{2} \end{gathered}$$
  Coordinates of Q are $(3,–\frac{9}{2},9)$ 
Equation of PN is  $\frac{x–1}{3}=\frac{y–2}{–176}=\frac{z–5}{–89}$