$\mathbf{The\;General \;Method}$\ Let $\boldsymbol{a_1}$, $\boldsymbol{a_2}$ and $\boldsymbol{a_3}$ be the crystallographic axes. Then the planes with <PERSON> indices $(hkl)$ intercept the crystallographic axes at $n\boldsymbol{a_1}/h,n\boldsymbol{a_2}/k$ and $n\boldsymbol{a_3}/l$, where $n$ is an integer. In general, the vectors $h\boldsymbol{a_1}+k\boldsymbol{a_2}+l\boldsymbol{a_3}$ is not perpendicular to the $(hkl)$ planes. This is why the general method of determination of the interplanar spacing requires reciprocal lattice basis $\boldsymbol{b_1},\boldsymbol{b_2}$ and $\boldsymbol{b_3}$, as the reciprocal lattice vectors $\boldsymbol{G}{hkl}=h\boldsymbol{b_1}+k\boldsymbol{b_2}+l\boldsymbol{b_3}$ is perpendicular to the family of $(hkl)$ planes. Expoliting this property, the $(hkl)$ interplanar spacing $d{hkl}$ can be found by projecting the vector joining the separation of two adjacent planes at the $\boldsymbol{a_1}$ axis onto the unit normal, that is $$d_{hkl}=\frac{\boldsymbol{a_1}}{h}\cdot \frac{\boldsymbol{G}{hkl}}{|\boldsymbol{G}{hkl}|}. \tag{1}\label{eq1}$$ Since $\boldsymbol{b_i}\cdot \boldsymbol{a_j}=2\pi\delta_{ij},$ Eq.\eqref{eq1} can be expressed as $$d_{hkl}=\frac{2\pi}{|\boldsymbol{G}{hkl}|}. 374
\tag{2}\label{eq2}$$\ $\mathit{Simple\, Cubic \,Lattice}$\ For simple cubic lattice, $|\boldsymbol{a_1}|=|\boldsymbol{a_2}|=|\boldsymbol{a_3}|=a.$ The reciprocal lattice vector is $\boldsymbol{G}{hkl}=\frac{2\pi}{a}(h\hat x+k\hat y+l\hat z),$ where $\hat x, \hat y$ and $\hat z$ are orthogonal unit vectors and by Eq.\eqref{eq2}, the interplanar spacing is $$d_{hkl}=\frac{a}{\sqrt{h^2+k^2+l^2}}. \tag{3}\label{eq3}$$
$\mathbf{Your\;Method}$\ Having presented the general formalism, now I attempt to use your reasoning to find the interplanar spacing for the simple cubic lattice. The nearest $(hkl)$ plane from the origin is found by setting $n=1,$ that is the plane which intercepts the crystallographic axes at $\boldsymbol{a_1}/h,\boldsymbol{a_2}/k$ and $\boldsymbol{a_3}/l$. Let the plane equation for this plane has the form $Ax/a+By/a+Cz/a=D.$ The x-interception is $a/h$, so we must have $D/A=1/h$. By similar reasoning, we get $D/B=1/k$ and $D/C=1/l$. So the plane equation has the form $$\frac{hx}{a}+\frac{ky}{a}+\frac{lz}{a}=1. \tag{4}\label{eq4}$$ From Eq.\eqref{eq4}, we see that the normal vector to the plane is $\frac{h\hat x}{a}+\frac{k\hat y}{a}+\frac{l\hat z}{a}.$ It can be shown that the adjacent plane to this plane should be $hx/a+ky/a+lz/a=0$ or $hx/a+ky/a+lz/a=2. 374