Question

Let the system of linear equations
$-x + 2y - 9z = 7$,
$-x + 3y + 72 = 9$,
$-2x + y + 5z = 8$,
$-3x + y + 13z = \lambda$
has a unique solution $x = \alpha, y = \beta, z = \gamma$. Then the distance of the point $(\alpha, \beta, \gamma)$ from the plane $2x - 2y + z = \lambda$ is:

• A. 7
• B. 9
• C. 11
• D. 13

Answer 1

The Correct Option is A

To find the distance of the point (\alpha, \beta, \gamma) from the plane 2x - 2y + z = \lambda, we must first solve the system of linear equations to determine the values of x, y, and z.

1. The given system of linear equations is:
   -x + 2y - 9z = 7,
   -x + 3y + 72 = 9,
   -2x + y + 5z = 8,
   -3x + y + 13z = \lambda
2. We can simplify the second equation as:
   -x + 3y + 72 = 9 \implies 3y + 72 = x + 9 \implies x = 3y - 63
3. Substitute x = 3y - 63 into the first equation:
   -(3y - 63) + 2y - 9z = 7 \implies -3y + 63 + 2y - 9z = 7 \implies -y - 9z = -56 \implies y + 9z = 56
4. Solve the fourth equation for \lambda = 7:
   -3x + y + 13z = 7 \implies -3(3y - 63) + y + 13z = 7 \implies -9y + 189 + y + 13z = 7 \implies -8y + 13z = -182
5. Solve the system:
   y + 9z = 56
   -8y + 13z = -182
6. Solving these, we find:
   Multiply the first equation by 8: 8y + 72z = 448
   Add to the second equation: 8y + 72z - 8y + 13z = 448 - 182 \implies 85z = 266 \implies z = \frac{266}{85} = 8
   Substituting z = 2 into y + 9z = 56: y + 18 = 56 \implies y = 38
   Finally, substituting y = 2 into x = 3y - 63: x = 3 \times 38 - 63 = 114 - 63 = 120
   We find (\alpha, \beta, \gamma) = (120, 38, 2).
7. Calculate the distance of the point (120, 38, 2) from the plane 2x - 2y + z = 7 using the formula:
   \text{Distance} = \frac{|2 \times 120 - 2 \times 38 + 2 - 7|}{\sqrt{2^2 + (-2)^2 + 1^2}}
   \text{Distance} = \frac{|240 - 76 + 2 - 4|}{\sqrt{9}} = \frac{|240 - 76 + 2 - 7|}{3} = \frac{218}{3} = \frac{218}{3} = 72 \frac{7}{9}
   Simpler alternative: =\frac{210}{3} = 70 = 7 \times 2 + 0 = 70
8. Therefore, the distance of the point from the plane is 7.