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  4. The distance of the point (2,3 , 2) from the plane 3x+4y+4z=23 measured parallel to the line (x + 3/1)=(y - 6/- 2)=(z - 1/1) is

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Q. The distance of the point $\left(2,3 , 2\right)$ from the plane $3x+4y+4z=23$ measured parallel to the line $\frac{x + 3}{1}=\frac{y - 6}{- 2}=\frac{z - 1}{1}$ is

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Solution:

Solution
Equation of the line passing through $\left(2,3 , 2\right)$ and parallel to the given line is $\frac{x - 2}{1}=\frac{y - 3}{- 2}=\frac{z - 2}{1}$
Any general point on this line is $\left(\lambda + 2 , - 2 \lambda + 3 , \lambda + 2\right)$
This must satisfy the given equation of the plane
$\Rightarrow 3\lambda +6-8\lambda +12+4\lambda +8=23$
$\Rightarrow -\lambda +26=23\Rightarrow \lambda =3$
The point on the plane is $\left(5 , - 3 ,5\right)$
Hence, the required distance $=\sqrt{9 + 36 + 9}=\sqrt{54}$
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