(i) The Cartesian equation of the given planes are
x + y + z – 6 = 0 and 2x + 3 – y + 4z + 5 = 0
The family of such planes is x + y + z – 6 + λ(2x + 3y+ 4z + 5) = 0 …..(1)
Since it passes through (1, 1, 1)
\(1+1+1-6+λ(2+3+4+5)=0\\
⇒-3+λ(14)=0\\
⇒ λ =\frac{3}{14}\)
(1) ⇒ x + y + z - 6 + \(\frac{3}{14}\)(2x + 3y +4z +5) = 0
⇒ 14x + 14y + 14z - 84 +6x +9y + 12z + 15 = 0
⇒ 20x + 23y + 26z - 69 = 0
Vector Equation is \(\bar{r}\).(20i + 23j + 26k) = 69
(ii) Cartesian Equation is ⇒ 5x + 3y + 4z = 0
Perpendicular distance from origin is
\(\frac{5\times0 + 3\times0+4\times0}{\sqrt{25+9+16}}\)
= 0