$(10$ points$)$ Find the distance from the line

$x=t$

$y=-t+1$

$z=2t+1$

to the line, which is the intersection between plane $x+y-z=1$ and plane $2x+z=3.($Hint: One can find the equation of the intersection line by solving the two equations of the planes together with the substitution $x=t.)$

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$6.(10$ points$)$ Find the distance from the line

$x=2+t$

$y=1+t$

$z=-\frac{1}{2}-\frac{1}{2}t$

to the plane $x+2y+6z=10.$

Please help me to solve the above question and provide a correcf step and answer.