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Q.
The distance of the point $(1,-5,9)$ from the plane $x-y+z=5$ measured along a straight line $x=y=z$ is
Solution:
Equation of $P Q$, is $\frac{x-1}{1}=\frac{x+5}{1}=\frac{z-9}{1}$
$x=\lambda+1 y=\lambda-5 \quad z=\lambda+9$ lies in
Then plane $x-y+z=5$
$\lambda+1-\lambda+5+\lambda+9=5$
$\lambda=-10$
$Q(-9,-15,-1)$ and $P(1,-5,9)$
$\Rightarrow |P Q|=\sqrt{10^{2}+10^{2}+10^{2}}=\sqrt{300}=10 \sqrt{3}$