(a) Cartesian equation is \(\frac{x-x_{1}} {a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\, \)
\(\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}\)
(b) Let
\(\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}\) = λ
x = 3λ + 1, y = λ + 2, z = 2λ + 3 are the general point of the line.
(c) Equation of the plane is 2x + 3y – z + 20….(1)
putting the values of x, y, z in (1) is
2(3λ + 1) – 3(λ + 2) – (2λ + 3) + 2 = 0
7λ + 7 = 0 ⇒ λ = -1
∴ Point of intersection is x = -3 + 1 = -2; y = -1 + 2 = 1, z = -2 + 3 = 1
ie; (-2, 1, 1)
(d) Distance =
\(\frac{2(1)+3(2)-3+2}{\sqrt{4+9+1}}\\=\frac{7}{\sqrt{14}}\)