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Q.
The distance of the plane $3 \,x+4 \,y+5 \,z+19=0$ from the point $(1,-1,1)$
measured along a line parallel to the line with direction ratios $2,3,1$ is
According to given information
Equation of lines is passing through $(1,-1,1)$ and
having DC's is $(2,3,1)$
$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{1}=r$
Here, $(2 \,r+1,3 \,r-1, r+1)$ lie on plane.
$\therefore $ These points satisfy the equation of plane.
$3(2 r+1)+4(3 r-1)+5(r+1)+19=0$
$\Rightarrow 6\, r+3+12 \,r-4+5\, r+5+19=0$
$\Rightarrow 23\, r+23=0 \Rightarrow r=-1$
So, point is $(-2+1,-3-1,-1+1)$ i.e. $(-1,-4,0)$
Now, required distance $=\sqrt{(-2)^{2}+(-3)^{2}+1^{2}}$
$=\sqrt{4+9+1}=\sqrt{14}$
