<s>[INST] Find the distance between the line and plane Find the distance between the line $x-z=3, x+2y+4z=6$ and the plane $3x+2y+2z=5$.
What I have done so far,
Found the vector by crossing $1,0-1$ and $1,2,4$ from the first line
The line is parallel to plane $\langle 2,-5,2\rangle \cdot \langle 3,2,2 \rangle = 0$
I think my next step is to find a point on Line 1 which satisfies both equations and then insert those values into the plane $3(x)+2(y)+2(z)=5$ and use the formula, $$
\frac{(3(x)+2(y)+2(z)-5)}{(3^2+2^2+2^2)}$$ to find the distance.
The values that I calculuate do not match the posted answer of $7/\sqrt{17}$
 [/INST] If $z=t,$  for the given line $x=z+3=t+3$ and
$x+2y+4z=6$ or, $t+3+2y+4(t)=6$ or, $2y=3-5t$
So, any point on the given line is $(t+3,t,\frac{3-5t}2)$
The distance of $(t+3,t,\frac{3-5t}2)$ from $3x+2y+2z=5$ is $$\left|\frac{3(t+3)+(3-5t)+2(t)-5}{\sqrt{3^2+2^2+2^2}}\right|=\frac 7{\sqrt {17}}$$
Had it not been a constant(which is due to the parallelism), we had to find $t$such that the distance is minimum.
</s>