At point P: x = -2, y = 6, z = 3. Hence

$\rho =\sqrt{{x}^{2}+{y}^{2}}=\sqrt{4+36}=6.32$

$\varphi ={\mathrm{tan}}^{-1}\frac{y}{x}={\mathrm{tan}}^{-1}\frac{6}{-2}={108.43}^{\circ}$

$r=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}=\sqrt{4+36+9}=7$

$\theta ={\mathrm{tan}}^{-1}=\frac{\sqrt{{x}^{2}+{y}^{2}}}{z}={\mathrm{tan}}^{-1}\frac{\sqrt{40}}{3}={64.42}^{\circ}$

Thus

$P(-2,6,3)=P(6.32,{108.43}^{\circ},3)=P(7,{64.62}^{\circ},{108.43}^{\circ})$

$\left[\begin{array}{c}{A}_{r}\\ {A}_{\theta}\\ {A}_{\varphi}\end{array}\right]=\left[\begin{array}{ccc}\mathrm{sin}\theta \mathrm{cos}\varphi & \mathrm{sin}\theta \mathrm{sin}\varphi & \mathrm{cos}\varphi \\ \mathrm{cos}\theta \mathrm{cos}\varphi & \mathrm{cos}\theta \mathrm{sin}\varphi & -\mathrm{sin}\theta \\ -\mathrm{sin}\theta & \mathrm{cos}\theta & 0\end{array}\right]\left[\begin{array}{c}y\\ x+z\\ 0\end{array}\right]$

or

${A}_{r}=y\mathrm{sin}\theta \mathrm{cos}\varphi +(x+z)\mathrm{sin}\theta \mathrm{sin}\varphi $

${A}_{\theta}=y\mathrm{cos}\theta \mathrm{cos}\varphi +(x+z)\mathrm{cos}\theta \mathrm{sin}\varphi $

${A}_{\varphi}=-y\mathrm{sin}\varphi +(x+z)\mathrm{cos}\varphi $

But $x=r\mathrm{sin}\theta \mathrm{cos}\varphi ,y=r\mathrm{sin}\theta \mathrm{sin}\varphi ,$ and $z=r\mathrm{cos}\theta $ . Substituting these yields

$A=({A}_{r},{A}_{\theta},{A}_{\varphi})$

$=r[{\mathrm{sin}}^{2}\theta \mathrm{cos}\varphi \mathrm{sin}\varphi +(\mathrm{sin}\theta \mathrm{cos}\varphi +\mathrm{cos}\theta )\mathrm{sin}\theta \mathrm{sin}\varphi ]{a}_{r}+r[\mathrm{sin}\theta \mathrm{cos}\theta \mathrm{sin}\varphi \mathrm{cos}\varphi +(\mathrm{sin}\theta \mathrm{cos}\varphi +\mathrm{cos}\theta )\mathrm{cos}\theta \mathrm{sin}\varphi ]{a}_{\theta}+r[-\mathrm{sin}\theta {\mathrm{sin}}^{2}\varphi +(\mathrm{sin}\theta \mathrm{cos}\varphi +\mathrm{cos}\theta )\mathrm{cos}\varphi ]{a}_{\varphi}$

At P

$r=7,\mathrm{tan}\varphi =\frac{6}{-2},\mathrm{tan}\theta =\frac{\sqrt{40}}{3}$

Hence, $\mathrm{cos}\varphi =\frac{-2}{\sqrt{40}},\mathrm{sin}\varphi =\frac{6}{\sqrt{40}},\mathrm{cos}\theta =\frac{3}{7},\mathrm{sin}\theta =\frac{\sqrt{40}}{7}$