Question

If point p (a,-8,4) and q(-3,-5,4) are
atdistace of 5 units. Find the value of a.

Answer 1

$\sqrt{[a+3]^{2}+[-8+5]^{2}+[4-4]^{2}   } =5\\\[a+3]^{2}+[-8+5]^{2}+[4-4]^{2} =25 \\\[a+3]^2 +[-3]^2  +0 =25\\\[a+3]^2 +9=25 \\\[a+3]^2 =16 \\a+3 = 4 or -4 \\a=4-3  or -4-3\\\\a = 1 or  -7$

a = 1  or  -7