If P represents radiation pressure, C represents speed of light and Q represents radiation energy striking a unit area per second, then non-zero integers x, y and z such that PˣQʸ Cᶻ is dimensionless, are

The correct option is x = 1, y = -1, z = 1.

To find the values of the non-zero integers $x$, $y$, and $z$ such that the expression $P^x{Q}^y{C}^z$ is dimensionless, we first need to determine the dimensional formula for each of the physical quantities involved.

1. Radiation Pressure ($P$):
Pressure is defined as force per unit area.
$\left[\text{Force}\right]=\left[{\text{MLT}}^{-2}\right]$
$\left[\text{Area}\right]=\left[{\text{L}}^2\right]$
Therefore, the dimensional formula for pressure $P$ is:
$\left[P\right]=\frac{\left[{\text{MLT}}^{-2}\right]}{\left[{\text{L}}^2\right]}$

2. Radiation Energy striking a unit area per second ($Q$):
This quantity represents energy per unit area per unit time.
$\left[\text{Energy}\right]=\left[{\text{ML}}^2{\text{T}}^{-2}\right]$
$\left[\text{Area}\right]=\left[{\text{L}}^2\right]$
$\left[\text{Time}\right]=\left[\text{T}\right]$
Therefore, the dimensional formula for $Q$ is:
$\left[Q\right]=\frac{\left[{\text{ML}}^2{\text{T}}^{-2}\right]}{\left[{\text{L}}^2\right]\left[\text{T}\right]}$

3. Speed of Light ($C$):
Speed is distance traveled per unit time.
$\left[C\right]=\left[{\text{LT}}^{-1}\right]$

Now, we set up the requirement for the expression $P^x{Q}^y{C}^z$ to be dimensionless:
$\left[P^x{Q}^y{C}^z\right]=\left[{\text{M}}^0{\text{L}}^0{\text{T}}^0\right]$

Substitute the dimensional formulas into the expression:
${\left[{\text{ML}}^{-1}{\text{T}}^{-2}\right]}^x{\left[{\text{MT}}^{-3}\right]}^y{\left[{\text{LT}}^{-1}\right]}^z=\left[{\text{M}}^0{\text{L}}^0{\text{T}}^0\right]$

Combining the exponents for each fundamental dimension:
$\left[{\text{M}}^{x+y}{\text{L}}^{-x+z}{\text{T}}^{-2x-3y-z}\right]=\left[{\text{M}}^0{\text{L}}^0{\text{T}}^0\right]$

By equating the powers on both sides, we get the following system of linear equations:
1) For M: $x+y=0\Rightarrow y=-x$
2) For L: $-x+z=0\Rightarrow z=x$
3) For T: $-2x-3y-z=0$

Substituting $y=-x$ and $z=x$ into the third equation:
$-2x-3(-x)-x=-2x+3x-x=0$
This equation is satisfied for any values of $x$, $y$, and $z$ as long as $y=-x$ and $z=x$.

Letting $x=1$, we find:
$y=-1$
$z=1$

This matches the correct option: x = 1, y = -1, z = 1.