Solution:

First, let's find the equations of the xy, yz, and zx planes.

Equation of the xy plane:
The xy plane is parallel to the z-axis and passes through the point (0,0,0). The equation of the xy plane can be written as z = 0.

Equation of the yz plane:
The yz plane is parallel to the x-axis and passes through the point (0,0,0). The equation of the yz plane can be written as x = 0.

Equation of the zx plane:
The zx plane is parallel to the y-axis and passes through the point (0,0,0). The equation of the zx plane can be written as y = 0.

Now, let's find the perpendicular distance from the point P(2,-3,5) to each of these planes.

Perpendicular distance from P to the xy plane (d1):
The equation of the xy plane is z = 0. The perpendicular distance from a point (x,y,z) to a plane with equation ax + by + cz + d = 0 is given by the formula:

d1 = |ax + by + cz + d| / sqrt(a^2 + b^2 + c^2)

Substituting the values, we have:

d1 = |0(2) + 0(-3) + 1(5) + 0| / sqrt(0^2 + 0^2 + 1^2)
= |5| / 1
= 5

Perpendicular distance from P to the yz plane (d2):
The equation of the yz plane is x = 0. Using the same formula as above, we have:

d2 = |0(0) + 1(-3) + 0(5) + 0| / sqrt(0^2 + 1^2 + 0^2)
= |-3| / 1
= 3

Perpendicular distance from P to the zx plane (d3):
The equation of the zx plane is y = 0. Using the same formula as above, we have:

d3 = |1(2) + 0(0) + 0(5) + 0| / sqrt(1^2 + 0^2 + 0^2)
= |2| / 1
= 2

Now, let's calculate the given expression:

d1^2 + 2d1^2 + 3d3^2 = 5^2 + 2(5^2) + 3(2^2)
= 25 + 2(25) + 3(4)
= 25 + 50 + 12
= 87

Therefore, the value of d1^2 + 2d1^2 + 3d3^2 is 87.

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