Answer
The greatest value of $x_p$ at which destructive interference occurs is $7.88 \mu \mathrm{m}$.
Work Step by Step
The path difference between the two light waves originating at $S_1$ and $S_2$ at interfering at P is,
$\Delta L=S_2P-S_1P$
Let $S_1P=x_p$ and $S_1S_2=d$, then from geometry, $S_2P=\sqrt{d^2+x_p^2}$. Then
$\Delta L=\sqrt{d^2+x_p^2}-x_p$
For destructive interference to occur at P, the path difference must be half integral multiple of the wavelength. I.e.,$\Delta L=\left(m+{1\over 2}\right)\lambda$
Hence,
$\sqrt{d^2+x_p^2}-x_p=\left(m+{1\over 2}\right)\lambda$
Rearranging and solving for $x_p$,
\begin{align*}
\sqrt{d^2+x_p^2}&=\left(m+{1\over 2}\right)\lambda+x_p\\
$d^2+x_p^2&=\left(\left(m+{1\over 2}\right)\lambda+x_p\right)^2\\
$d^2+x_p^2&=\left(m+{1\over 2}\right)^2\lambda^2+x_p^2+2x_p(m+{1\over 2})\lambda\\
d^2&={(2m+1)^2\over 4}\lambda^2+2x_p\left(m+{1\over 2}\right)\lambda\\
2x_p\left(m+{1\over 2}\right)\lambda&=d^2-{(2m+1)^2\over 4}\lambda^2\\
x_p&={d^2\over \lambda (2m+1)}-{2m+1\over 4}\lambda
\end{align*}
To obtain the maximum value for $x_p$, we set $m=0$ (first dark fringe),
$x_p={d^2\over\lambda}-{\lambda\over 4}$
Plugging in the given values $\lambda=900\times 10^{-9}\mathrm{m}, d=2.70\times 10^{-6}\mathrm{m}$, we get,
\begin{align*}
x_p&={7.29\times 10^{-12}\over 900\times 10^{-9}}-{900\times 10^{-9}\over 4}\\
&=0.0081\times 10^{-3}-225\times 10^-9\\
&=8.1\times 10^{-6}-0.225\times 10^{-6}\\
&=7.875\times 10^{-6}\\
&=7.88\mu \mathrm{m}
\end{align*}
Hence, the greatest value of $x_p$ at which destructive interference occurs is $7.88 \mu \mathrm{m}$.
