(a) since, distance of a point P(a,b,c) from y − axis $=\sqrt{{\mathrm{a}}^2+{\mathrm{z}}^2}$  so , here distance of point P(−3,−1,6) from$\mathrm{y}-\mathrm{axis}=\sqrt{(-3{)}^2+(6{)}^2}$ 

$=\sqrt{9+36}=\sqrt{45}=3\sqrt{5}\text{ units }$

(b) given point is P(−3,−1,6) If we draw a perpendicular from the point P(−3,−1,6) on xz−plane then the foot of the perpendicular is (−3,0,6)

Now the distance of the point P(−3,−1,6) from the xz−plane is the distance between the ponts P(−3,−1,6) and (−3,0,6) and that $\sqrt{(-3+3{)}^2+(-1-0{)}^2+(6-6{)}^2}=\sqrt{0+1+0}=1\text{ unit }$

(c) If a point has coordinates (x,y,z) , then the image of this point in the XY−plane is given by (x,y,−z) so, here Image of P(−3,−1,6) with respect to XY−plane is given by (−3,−1,−6) (d) The x−coordinate, y−coordinate and z−coordinate of point (−3,−1,6) are negative negative and positive respectively.

Therefore this point lies in octant III