The distance of the plane 3x+4y+5z+19=0 from the point (1,-1,1) measured along a line parallel to the line with direction ratios 2,3,1 is

# The distance of the plane $3x+4y+5z+19=0$ from the point (1,-1,1) measured along a line parallel to the line with direction ratios 2,3,1 is

1. A

$\frac{23}{5\sqrt{2}}$

2. B

$\frac{\sqrt{71}}{5\sqrt{2}}$

3. C

$\sqrt{14}$

4. D

$\sqrt{23}$

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### Solution:

Line passing through (1,-1,1) is $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{1}=t$

$P\left(2t+1,3t-1,t+1\right)$ lies on $3x+4y+5z+19=0$

$t=-1⇒P\left(-1,-4,0\right)Q\left(1,-1,1\right)$

$PQ=\sqrt{14}$

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