QJEEAdv15Ma1fd4: JEE Advanced 2015 Mathematics - Class 12 JEEAdv15Ma1fd4 Solution: In consider the planes and Let be the plane, different from and …

$\begin{aligned}& \text { Given, } \mathrm{P}_1: y=0 \quad \text{... (i)}\\& \mathrm{P}_2: x+z-1=0 \quad \text{... (ii)}\end{aligned}$

Equation of plane $P_3$ passing through the intersection of plane $P_1$ and $P_2$ is given by

$\mathrm{P}_3: x+z-1+\lambda y=0 \quad \text{... (iii)}$

$\because$ Distance of the point $(0,1,0)$ from $P_3$ is 1.

$\begin{aligned}& \therefore \quad\left|\frac{0+\lambda+0-1}{\sqrt{1^2+\lambda^2+1^2}}\right|=1 \\& \Rightarrow \quad\left|\frac{\lambda-1}{\sqrt{2+\lambda^2}}\right|=1 \\& \Rightarrow \quad(\lambda-1)^2=1\left(\sqrt{2+\lambda^2}\right)^2 \\& \Rightarrow \quad \lambda^2+1-2 \lambda=\lambda^2+2 \\& \Rightarrow \quad \lambda=-\frac{1}{2} \\\end{aligned}$

Put the value of $\lambda$ in equation (iii), we get

$\begin{aligned}& \mathrm{P}_3: \quad x+z-1-\frac{1}{2} y=0 \\& \Rightarrow \quad 2 x-y+2 z-2=0\end{aligned}$

$\because$ Distance of a point $(\alpha, \beta, \gamma)$ from $\mathrm{P}_3$ is 2

$\begin{array}{ll}\therefore & \left|\frac{2 \alpha-\beta+2 \gamma-2}{\sqrt{2^2+1^2+2^2}}\right|=2 \\\Rightarrow & \left|\frac{2 \alpha-\beta+2 \gamma-2}{3}\right|=2 \\\Rightarrow & 2 \alpha-\beta+2 \gamma-2= \pm 6 \\\Rightarrow & 2 \alpha-\beta+2 \gamma=8 \text { or }-4 \\\Rightarrow & 2 \alpha-\beta+2 \gamma-8=0 \text { or } 2 \alpha-\beta+2 \gamma+4=0\end{array}$

Hint :

(i) Equation of plane passing through the intersection of two plane $P_1$ and $P_2$ is given by $\mathrm{P}_1+\lambda \mathrm{P}_2=0$ ; where $\lambda$ is a constant.

(ii) Distance of a point $\left(x_1 y_1 z_1\right)$ from plane $a x +b y+c z+d=0$ is $\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right|$