Answer
The lengths are $|PQ|=6$, $|PR|=6$, and $|QR|=2\sqrt{10}.$
The triangle is isosceles but is not a right triangle.
Work Step by Step
To find the lengths of the legs of the triangle, we must find the distance between each pair of points.
$$|PQ|=\sqrt{(7-3)^2+(0+2)^2+(1+3)^2}
\\=\sqrt{(4)^2+(2)^2+(4)^2} \\=\sqrt{16+4+16}\\=\sqrt{36}\\=6.$$
$$|PR|=\sqrt{(1-3)^2+(2+2)^2+(1+3)^2}
\\=\sqrt{(-2)^2+(4)^2+(4)^2} \\=\sqrt{4+16+16}\\=\sqrt{36}\\=6.$$
$$|QR|=\sqrt{(1-7)^2+(2-0)^2+(1-1)^2}
\\=\sqrt{(-6)^2+(2)^2+(0)^2} \\=\sqrt{36+4+0}\\=\sqrt{40}\\=\sqrt{4(10)}\\=2\sqrt{10}.$$
Now, if a triangle has two sides of equal length, then it is isosceles. So since $|PQ|=|PR|=6$, our triangle is isosceles.
And for a triangle to be a right triangle, the square of the longest side must equal the sum of the squares of the other two sides. The longest side of our triangle is $|PQ|=2\sqrt{10}$, but $$|PQ|^2+|PR|^2=36+36=72\neq 40=|QR|^2.$$
So our triangle is not a right triangle.
