$\because$ Line is at constant distance from both the planes

$\therefore$ Line will be parallel to both planes 

$\therefore$ perpendicular to their normals

Normal will be  $\overrightarrow{{\mathrm{n}}_1}=\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\mathrm{and}\overrightarrow{{\mathrm{n}}_2}=2\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\therefore \overrightarrow{{\mathrm{n}}_1}\times \overrightarrow{{\mathrm{n}}_2}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& 2& -1\\ 2& -1& 1\end{array}\right|=\hat{\mathrm{i}}-3\hat{\mathrm{j}}-5\hat{\mathrm{k}}$

$\therefore$ Equation of line will be   = $\frac{\mathrm{x}-0}{1}=\frac{\mathrm{y}-0}{-3}=\frac{\mathrm{z}-0}{-5}$

Let foot of perpendicular from (0, 0, 0) on plane P₁: x + 2y – z + 1 = 0 be (α, β, γ)

 $\therefore \frac{\mathrm{\alpha }-0}{1}=\frac{\mathrm{\beta }-0}{2}=\frac{\mathrm{\gamma }-0}{-1}=\frac{-(0+0+0+1)}{1^2+2^2+{(-1)}^2}=\frac{-1}{6}$

 $\therefore \mathrm{\alpha }=\frac{-1}{6},\mathrm{\beta }=\frac{–2}{6}\mathrm{and}\mathrm{\gamma }=\frac{1}{6}$

Locus of feet of perpendicular drawn from line upon plane will be a parallel line passing through $(\frac{-1}{3},\frac{2}{3},\frac{1}{3})$ = (α, β, γ)

 $\therefore$ ne will be   Li$\frac{\mathrm{x}+\frac{1}{6}}{1}=\frac{\mathrm{y}+\frac{2}{6}}{-3}=\frac{\mathrm{z}-\frac{1}{6}}{-5}=\mathrm{\lambda }$

(A)        If x = 0

 $\therefore \mathrm{y}+\frac{2}{6}=\frac{–3}{6}\Rightarrow \mathrm{y}=\frac{–5}{6}\mathrm{and}–\frac{1}{6}=\frac{–5}{6}\Rightarrow \frac{–4}{6}=\frac{–2}{3}$

$\therefore$ Point is $(0,\frac{-5}{6},\frac{-2}{3})$ .

(B)        $\text{If\hspace{0.17em}x=}\frac{-1}{6}\mathrm{\hspace{0.17em}},\mathrm{then}\mathrm{\hspace{0.17em}}=\frac{–2}{6}\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{\hspace{0.17em}}\mathrm{z}=\frac{1}{6}$

(C) & (D)

If y = 0, then   $\frac{\mathrm{x}+\frac{1}{6}}{1}=\frac{0+\frac{2}{6}}{-3}=\frac{\mathrm{z}-\frac{1}{6}}{-5}$

 $\Rightarrow \mathrm{x}+\frac{1}{6}=\frac{–1}{9}\Rightarrow \mathrm{x}=-\frac{1}{6}-\frac{1}{9}=\frac{-3-2}{18}=\frac{-5}{18}$

 $\Rightarrow \frac{1}{6}=\frac{5}{9}\Rightarrow z=\frac{5}{9}+\frac{1}{6}=\frac{13}{18}.$

⇒ Answer is (A) & (B).