Let A (a, b, c) is foot of perpendicular of point P (-1, -2, 3)
Given plane is x - y + 3z = 2.
So, PA will be normal to the given plane, so direction ratios of PA will be proportional to (1, -1, 3).
∵ PA passes from (-1, -2, 3) and have direction ratios (1, -1, 3).
∴ Equation of line
PA====r( say
) So, point A (a, b, c) in the form of r is
(r−1,−r−2,3r+3) Since A (a, b, c), lies on the given plane, so,Plane
x−y+3z=2⇒1.(r−1)–1.(−r−2) +3(r+3)–2=0 ∴r= So, foot of perpendicular
A(r−1,−r−2,3r+3) will be
(,,)