If the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ is contained in the plane $px+qy-z=9$, then the value of $p^{2}+q^{2}$ is:

Question

Accepted Answer

To determine why Option 4 is the correct answer, we need to analyze the given line and plane equations and find the value of $ p^2 + q^2 $.

The line is given in symmetric form:
$$
\frac{x-3}{2} = \frac{y+2}{-1} = \frac{z+4}{3}
$$
This can be parameterized as:
$$
x = 3 + 2t, \quad y = -2 - t, \quad z = -4 + 3t
$$
where $ t $ is a parameter.

The plane equation is:
$$
px + qy - z = 9
$$

For the line to be contained in the plane, every point $(x, y, z)$ on the line must satisfy the plane equation. Substituting the parameterized coordinates into the plane equation gives:
$$
p(3 + 2t) + q(-2 - t) - (-4 + 3t) = 9
$$

Simplifying this, we have:
$$
3p + 2pt - 2q - qt + 4 - 3t = 9
$$

Rearranging terms, we get:
$$
(2p - q - 3)t + (3p - 2q + 4) = 9
$$

For this equation to hold for all $ t $, both the coefficient of $ t $ and the constant term must independently equal zero. Thus, we have the system of equations:
1. $ 2p - q - 3 = 0 $
2. $ 3p - 2q + 4 = 9 $

Solving these equations:

From equation 1:
$$
2p - q = 3 \quad \text{(Equation A)}
$$

From equation 2:
$$
3p - 2q = 5 \quad \text{(Equation B)}
$$

We can solve these equations simultaneously. Multiply Equation A by 2:
$$
4p - 2q = 6
$$

Subtract Equation B from this result:
$$
(4p - 2q) - (3p - 2q) = 6 - 5
$$
$$
p = 1
$$

Substitute $ p = 1 $ back into Equation A:
$$
2(1) - q = 3
$$
$$
2 - q = 3
$$
$$
q = -1
$$

Now, calculate $ p^2 + q^2 $:
$$
p^2 + q^2 = 1^2 + (-1)^2 = 1 + 1 = 2
$$

Thus, the value of $ p^2 + q^2 $ is indeed 2, confirming that Option 4 is correct.

**Explanation of Incorrect Options:**

- **Option 1 (26):** This would imply much larger values for $ p $ and $ q $, which do not satisfy the derived equations.
- **Option 2 (18):** This would also imply incorrect values for $ p $ and $ q $ that do not satisfy the conditions for the line to lie in the plane.
- **Option 3 (5):** This suggests values for $ p $ and $ q $ that do not solve the system of equations derived from the line and plane conditions.

Therefore, the correct answer is Option 4, with $ p^2 + q^2 = 2 $.