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Mathematics
Mathematics
Problem
10. Find the values of
p
p
p
so that the lines
1
−
x
3
=
7
y
−
14
2
p
=
z
−
3
2
\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}
3
1
−
x
=
2
p
7
y
−
14
=
2
z
−
3
and
7
−
7
x
3
p
=
y
−
5
1
=
6
−
z
5
\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}
3
p
7
−
7
x
=
1
y
−
5
=
5
6
−
z
are at right angles.
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