The distance of the point (-1, -5, -10) from the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=5$ will be

Q: The distance of the point (-1, -5, -10) from the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=5$ will be

To determine the distance of the point $(-1, -5, -10)$ from the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=5$, we need to follow these steps: 1. **Find the Point of Intersection:** The line is given in the symmetric form: \[ \frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} = t \] From this, we can express $x$, $y$, and $z$ in terms of the parameter $t$: \[ x = 3t + 2, \quad y = 4t - 1, \quad z = 12t + 2 \] Substitute these into the plane equation $x - y + z = 5$: \[ (3t + 2) - (4t - 1) + (12t + 2) = 5 \] Simplify: \[ 3t + 2 - 4t + 1 + 12t + 2 = 5 \] \[ 11t + 5 = 5 \] \[ 11t = 0 \quad \Rightarrow \quad t = 0 \] Substitute $t = 0$ back into the parametric equations: \[ x = 3(0) + 2 = 2, \quad y = 4(0) - 1 = -1, \quad z = 12(0) + 2 = 2 \] Thus, the point of intersection is $(2, -1, 2)$. 2. **Calculate the Distance:** The distance $d$ between the points $(-1, -5, -10)$ and $(2, -1, 2)$ is given by the distance formula: \[ d = \sqrt{(2 - (-1))^2 + (-1 - (-5))^2 + (2 - (-10))^2} \] Simplify: \[ d = \sqrt{(2 + 1)^2 + (-1 + 5)^2 + (2 + 10)^2} \] \[ d = \sqrt{3^2 + 4^2 + 12^2} \] \[ d = \sqrt{9 + 16 + 144} \] \[ d = \sqrt{169} \] \[ d = 13 \] Therefore, the distance is 13, which corresponds to Option 4. 3. **Explanation of Incorrect Options:** - **Option 1 (10):** This option is incorrect because the calculated distance using the distance formula does not yield 10. - **Option 2 (8):** This option is incorrect for the same reason; the distance calculation does not result in 8. - **Option 3 (21):** This option is incorrect as the distance calculation clearly shows the result is not 21. In conclusion, the correct answer is Option 4, as the calculated distance between the given points is indeed 13.

Question

Accepted Answer

To determine the distance of the point $(-1, -5, -10)$ from the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=5$, we need to follow these steps:

1. **Find the Point of Intersection:**

The line is given in the symmetric form:
   $$
   \frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} = t
   $$
   From this, we can express $x$, $y$, and $z$ in terms of the parameter $t$:
   $$
   x = 3t + 2, \quad y = 4t - 1, \quad z = 12t + 2
   $$

Substitute these into the plane equation $x - y + z = 5$:
   $$
   (3t + 2) - (4t - 1) + (12t + 2) = 5
   $$
   Simplify:
   $$
   3t + 2 - 4t + 1 + 12t + 2 = 5
   $$
   $$
   11t + 5 = 5
   $$
   $$
   11t = 0 \quad \Rightarrow \quad t = 0
   $$

Substitute $t = 0$ back into the parametric equations:
   $$
   x = 3(0) + 2 = 2, \quad y = 4(0) - 1 = -1, \quad z = 12(0) + 2 = 2
   $$

Thus, the point of intersection is $(2, -1, 2)$.

2. **Calculate the Distance:**

The distance $d$ between the points $(-1, -5, -10)$ and $(2, -1, 2)$ is given by the distance formula:
   $$
   d = \sqrt{(2 - (-1))^2 + (-1 - (-5))^2 + (2 - (-10))^2}
   $$
   Simplify:
   $$
   d = \sqrt{(2 + 1)^2 + (-1 + 5)^2 + (2 + 10)^2}
   $$
   $$
   d = \sqrt{3^2 + 4^2 + 12^2}
   $$
   $$
   d = \sqrt{9 + 16 + 144}
   $$
   $$
   d = \sqrt{169}
   $$
   $$
   d = 13
   $$

Therefore, the distance is 13, which corresponds to Option 4.

3. **Explanation of Incorrect Options:**

- **Option 1 (10):** This option is incorrect because the calculated distance using the distance formula does not yield 10.
   - **Option 2 (8):** This option is incorrect for the same reason; the distance calculation does not result in 8.
   - **Option 3 (21):** This option is incorrect as the distance calculation clearly shows the result is not 21.

In conclusion, the correct answer is Option 4, as the calculated distance between the given points is indeed 13.