The perpendicular distance of the point \[(x_1,y_1,z_1)\] from the plane \[lx+my+nz+d=0\] is \[ |\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}|\]
Find the distance of the point (2,-1,5) from the plane \[ x-2y+5z+12=0\]
\[ \frac{2-2(-1)+5\times 5+12}{\sqrt{1^2+(-2)^2+5^2}}\]=\[ \frac{2+2+25+12}{\sqrt{30}}\]=\[ \frac{41}{\sqrt{30}}\]
