Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

For (P): The plane passing through $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ has the intercept form $\frac{x}{1}+\frac{y}{1}+\frac{z}{1}=1$, which simplifies to $x+y+z=1$. So, (P) matches (1).

For (Q): The perpendicular distance of a point $(x_1,y_1,z_1)$ from the plane $Ax+By+Cz+D=0$ is given by $D=\frac{|A{x}_1+B{y}_1+C{z}_1+D|}{\sqrt{A^2+B^2+C^2}}$. For point $(1,2,3)$ and plane $x-2y+2z-1=0$, the distance is $D=\frac{|1(1)-2(2)+2(3)-1|}{\sqrt{1^2+(-2{)}^2+2^2}}=\frac{|1-4+6-1|}{\sqrt{1+4+4}}=\frac{|2|}{\sqrt{9}}=\frac{2}{3}$. So, (Q) matches (2).

For (R): The cosine of the angle $\theta$ between two planes with normal vectors $\overrightarrow{n_1}$ and $\overrightarrow{n_2}$ is given by $\cos \theta =\frac{|\overrightarrow{n_1}\cdot \overrightarrow{n_2}|}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}$. For $x+y+z=1$, $\overrightarrow{n_1}=(1,1,1)$. For $2x-y+2z=3$, $\overrightarrow{n_2}=(2,-1,2)$.$\overrightarrow{n_1}\cdot \overrightarrow{n_2}=(1)(2)+(1)(-1)+(1)(2)=2-1+2=3$.$|\overrightarrow{n_1}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$.$|\overrightarrow{n_2}|=\sqrt{2^2+(-1{)}^2+2^2}=\sqrt{4+1+4}=\sqrt{9}=3$.$\cos \theta =\frac{|3|}{\sqrt{3}\cdot 3}=\frac{1}{\sqrt{3}}$. So, (R) matches (4).

For (S): A plane parallel to $x+2y+3z=0$ has an equation of the form $x+2y+3z=k$. Since it passes through $(1,1,1)$, we substitute these coordinates: $1+2(1)+3(1)=k⟹k=6$. Thus, the equation of the plane is $x+2y+3z=6$. So, (S) matches (3).