The point on the plane $2x-2y+4z+5=0$ that is nearer to $\left(1, \frac{3}{2}, 2\right)$ is

Let $P$ be the point nearer to $\left(1, \frac{3}{2}, 2\right)$ on plane $2x-2y+4z+5=0$

Direction ratios of the normal to the plane $\left(2,-2,4\right)$

$\therefore P = \left(1,\frac{3}{2}, 2\right) + c\left(2, -2, 4\right)$

As point $P$ lies on the plane $2x-2y+4z+5=0$.

$\Rightarrow 2\left(1+2c\right) - 2\left(\frac{3}{2}-2c\right) +4\left(2+4c\right) +5=0$

$\Rightarrow 2+4c-3+4c+8+16c+5=0$

$\Rightarrow c=-\frac{1}{2}$

$\therefore P = \left(1,\frac{3}{2}, 2\right) + \left(-\frac{1}{2}\right)\left(2, -2, 4\right)$

$\Rightarrow P=\left(0, \frac{5}{2}, 0\right)$