Given equation: 2x + 3y - 7 = 0

Finding the mean and mean deviation of x:
- To find the mean of x, we need to first isolate x in the equation.
2x + 3y - 7 = 0
2x = 7 - 3y
x = (7 - 3y)/2

- Now we can find the mean of x by taking the average of all possible values of x.
Mean of x = (1/∞)∫(7-3y)/2 dy from -∞ to +∞
= (1/∞)[(7y/2) - (3y^2/4)] from -∞ to +∞
= (1/∞)[(+∞) - (-∞)]
= 1

- To find the mean deviation of x about its mean, we need to find the deviation of each value of x from the mean, take the absolute value of each deviation, add them up, and divide by the number of values.
Mean deviation of x = (1/∞)∫|x-1| dy from -∞ to +∞
= (1/∞)∫|(7-3y)/2 - 1| dy from -∞ to +∞
= (1/∞)∫|3y/2 - 9/2| dy from -∞ to +∞
= (1/∞)∫(3y/2 - 9/2) dy from -∞ to 6
+ (1/∞)∫(-3y/2 + 9/2) dy from 6 to +∞
= (1/∞)[(3/4)*(6-(-∞))^2 - (3/4)*(0-(-∞))^2]
+ (1/∞)[(-3/4)*((+∞)-6)^2 - (-3/4)*((+∞)-0)^2]
= 0.3

Finding the coefficient of mean deviation of y about its mean:
- We know that the mean deviation of y about its mean is given by:
Mean deviation of y = (1/∞)∫|y-μ| dx from -∞ to +∞
where μ is the mean of y.

- To find μ, we need to isolate y in the equation.
2x + 3y - 7 = 0
3y = 7 - 2x
y = (7 - 2x)/3

- Now we can find μ by taking the average of all possible values of y.
μ = (1/∞)∫(7-2x)/3 dx from -∞ to +∞
= (1/∞)[(-7/6)*x + (1/3)*(x^2/2)] from -∞ to +∞
= 0

- Therefore, the coefficient of mean deviation of y about its mean is 0, since the mean deviation of y about its mean is independent of the value of μ.

50