Calculate the moment about a point (5; 4; 3) ft. caused by the forces

P = 3^i - 2^j - 4^k lb. and Q = 5^i + 7^j - 2^k lb. acting at the points
(7; 2; 2) ft. and (3; 2; 5) ft., respectively.

What I did was this;

vector r1= (7,2,2) - (5,4,3) = 2i-2j-5k

vector r2= (3,2,5) - (5,4,3) = -8i-6j+2k

Vector M1= (2i-2j-5k) X P = -7i-7j+2k
Magnitude= 7.55 ft-lbs

Vector M2= (-8i-6j+2k) X Q = -2i-6j+85k
Magnitude= 86.23 ft-lbs

Moment about the given point = (7.55) + (86.23) = 93.78 ft-lbs

I'm unsure if this is right

Your calculations are mostly correct, but there are a few small errors. Let's go through the steps to calculate the moment about the given point (5, 4, 3) ft caused by the forces P and Q.

1. Determine the position vectors from each force to the given point.
- Position vector r1 = (7, 2, 2) - (5, 4, 3) = 2i - 2j - k
- Position vector r2 = (3, 2, 5) - (5, 4, 3) = -2i - 2j + 2k

2. Calculate the cross product of each position vector with the corresponding force vector.
- Moment caused by force P:
M1 = r1 × P = (2i - 2j - k) × (3i - 2j - 4k)
= (2*(-4) - (-2)*(-4))i - (2*(-1) - 3*(-4))j + (2*(-2) - 3*(2))k
= -6i + 10j - 10k

- Moment caused by force Q:
M2 = r2 × Q = (-2i - 2j + 2k) × (5i + 7j - 2k)
= (-2*(-2) - 2*(-2))i - (-2*(5) - 2*(-2))j + (-2*(7) - (-2)*(-2))k
= 2i + 6j + 18k

3. Calculate the magnitudes of the moments using the formula |M| = √(Mx^2 + My^2 + Mz^2).
- Magnitude of M1 = √((-6)^2 + 10^2 + (-10)^2) ≈ 14.73 ft-lbs
- Magnitude of M2 = √(2^2 + 6^2 + 18^2) ≈ 19.62 ft-lbs

4. Find the total moment about the given point by summing the magnitudes of the moments.
Total moment = |M1| + |M2| ≈ 14.73 + 19.62 ≈ 34.35 ft-lbs

So, the correct moment about the point (5, 4, 3) ft caused by the forces P and Q is approximately 34.35 ft-lbs.