Question

The equation of a plane passing through the line of intersection of the planes $x + 2y + 3z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{\sqrt 3}$ from the point $(3,1, -1)$ is

• A. $5x-11y+z=17$
• B. $\sqrt {2x}+y = 3\sqrt 2-1$
• C. $ x+ y + z = \sqrt 3$
• D. $x -\sqrt {2y} = 1-\sqrt 2$

Answer 1

Answer: (A)

PLAN
(i) Equation of plane through intersection of two planes,
$ i.e, (a_1 x + b_1 y +c_1 z +d_1)+ \lambda $
$ (a_2 x +b_2 y +c_2 z +d_2) = 0$
(ii) Distance of a point $(x_1,y_1,z_1)$ from
$ ax+by+cz+d = 0$
$ =\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$
Equation of plane passing through intersection of two
planes $x + 2y + 3z = 2$ and $x - y + z = 3$ is
$ (x + 2y + 3z - 2) + \lambda (x - y + z - 3) = 0$
$\Rightarrow (1+ \lambda) x + (2+ \lambda) y +(3+ \lambda)z - (2+3\lambda) = 0$
whose distance from $(3,1, -1)$ is $\frac{2}{\sqrt 3}.$
$\Rightarrow \frac{|3\, (1+ \lambda) +1. (2 - \lambda) -1\, (3+ \lambda) - (2+3\lambda)| }{\sqrt {(1+ \lambda)^2 + (2 - \lambda)^2 + (3+ \lambda)^2}} = \frac{2}{\sqrt 3}$
$\Rightarrow $ $ \frac{| -2 \lambda\, | }{\sqrt {3\lambda^2 + 4 \lambda^2 +14}} = \frac{2}{\sqrt 3}$
$\Rightarrow 3\lambda^2=3\lambda^2+4\lambda+14 $
$\Rightarrow \lambda= -\frac{7}{2} $
$\therefore \bigg( 1 -\frac{7}{2}\bigg)x +\bigg( 2 -\frac{7}{2}\bigg)y+\bigg( 3 -\frac{7}{2}\bigg)z-\bigg( 2 -\frac{21}{2}\bigg) = 0 $
$\Rightarrow $ $-\frac{5x}{2}+\frac{11}{2}y-\frac{1}{2}z+\frac{17}{2}=0 $
or $ {5x} -{11} y +z+ 17 =0 $