Find the coordinates of the foot of perpendicular and the length of the perpendicular drawn from the point P(5,4,2) to the line →r=−^i+3^j+^k+μ(2^i+3^j−^k). Also find the image of P in this line.

Any point on the line can be written in parametric form as (2λ−1, 3λ+3, −λ+1). Assuming this as the foot of perpendicular from (5,4,2), we can equate the dot product of this vector and the line direction to zero.
∴((2λ−1)^i+(3λ+3)^j+(−λ+1)^k) ⋅ (2^i+3^j−^k)=0
⇒(2λ−1)×2+(3λ+3)×3+(−λ+1)×(−1)=0
⇒4λ−2+9λ+9+λ−1=0
⇒14λ+6=0
⇒λ=−3/7
The coordinates of the point are thus (2(−3/7)−1, 3(−3/7)+3, −(−3/7)+1) = (-13/7, 12/7, 10/7)
The length of the perpendicular can be found out by
√((5 + 13/7)²+(4 - 12/7)²+(2 - 10/7)²) = √((48/7)²+(16/7)²+(4/7)²) = √(2304+256+16)/49 = √(2576)/7 = 4√161/7
The foot of perpendicular would be the midpoint of P and the image of P in the line.
∴(5^i+4^j+2^k)+(x^i+y^j+z^k)=2×((-13/7)^i+(12/7)^j+(10/7)^k)
⇒x = 2(-13/7)-5 = -51/7, y = 2(12/7)-4 = 8/7, z = 2(10/7)-2 = 6/7
The image of point P is thus (−51/7,8/7,6/7)