Let P is a point on the plane $ax+by+cz=d.$ A point $Q(\alpha ,\beta ,\gamma )$ is taken on the line $OP$ such that $OP.OQ=d^2,$ then $\frac{2d(a\alpha +b\beta +c\gamma )}{{\alpha }^2+{\beta }^2+{\gamma }^2}$ is equal to (where O is origin)

Let $P(x_1,y_1,z_1)$ be a point on $ax+by+cz=d.$ Then $a{x}_1+b{y}_1+c{z}_1=d$

Let $OP=r.$ Then direction cosines of $OP$ are $\frac{x_1}{r},\frac{y_1}{r},\frac{z_1}{r}$

Equation of line OP is $\frac{x-0}{\frac{x_1}{r}}=\frac{y-0}{\frac{y_1}{r}}=\frac{z-0}{\frac{z_1}{r}}=\lambda$

Let $Q(\alpha ,\beta ,\gamma )$ be a point on OP such then $OQ=\lambda$ Then coordinates of $Q(\frac{x_1\lambda }{r},\frac{y_1\lambda }{r},\frac{z_1\lambda }{r}).$

But coordinate of $Q$ be $(\alpha ,\beta ,\gamma )$

$x_1=\frac{\alpha r}{\lambda },y_1=\frac{\beta r}{\lambda },z_1=\frac{\gamma r}{\lambda }\Rightarrow a{x}_1+b{y}_1+c{z}_1=d$

$\frac{r}{\lambda }(a\alpha +b\beta +c\gamma )=d\dots \mathrm{..}\text{ (ii)}$

$\text{OP.OQ }=d^2$

$\lambda =d^2$

$\frac{{\lambda }^2\text{ d}}{(a\alpha +b\beta +c\gamma )}=d^2$