Originally Posted by

**cellae** I was given:

**(a) Find the equation of the plane p which passes through the three points**

(A 1,0,1), B(2,−1,1) .and C(0,3,2) .

I got n = AB x AC

AB = (2 - 1)i + (-1 - 0)j + (1 - 1)k

= i - j

AC = (0 - 1)i + (3 - 0)j + (2 - 1)k

+ -i + 3j + k

And AB x AC

n = -i - j + 2k

(b) Find a scalar parametric form of the equation for the line which passes through the point D(−1,1,1) and which is perpendicular to the plane p.

I got r = r_{0} + tv

= (-1,1,1) + t (-1,-1,2)

= (-1,1,1) + (-t - t + 2t)

= (-1-t, 1-t, 1+2t)

so:

x = -1-t

y= 1-t

z= 1 + 2t

I'm confused what Q's c and d are asking...

**(c) Let E be the point where the line intersects the plane p. Find, in the scalar parametric equation for the line, the value of the parameter which corresponds to the point E and hence find the co-ordinates of this point.**

(d) What is the closest distance of the point D from the plane p?

Thank you !