Find the orthocentre of the triangle the equation of whose sides are x+y=1, 2x+3y=6, 4x-y+4=0

Let equation of
AB be  x+y-1 =0---(1)
BC be 2x+3y-6= 0 ---(2)
and AC be  4x-y+4=0 ---(3)

Solving (1) and (2) B = (- 3, 4 )
Solving (1) and (3) A =(-3/5, 8/5)

Equation of BC is  2x+3y=6

Altitude AD is perpendicular to BC, 

Therefore Equation of AD is x + y + k = 0

AD is passing through A (-3/5, 8/5)
⇒ (-3/5)+(8/5)+k=0

⇒k = -1 

∴ Equation if AD is x + y -1 = 0 ----(4)

Altitude BE is perpendicular to AC.

⇒Let the equation of DE be x – 2y = k

BE is passing through D  (- 3, 4 )
⇒-3-8=k

⇒ k = -11

Equation of BE is x – 2y = -11-----(5)

Solving (4) and (5), the point of intersection is  (-3, 4)
Therefore the orthocenter of the triangle is  (-3, 4)