If the foot of the perpendicular from the point $\mathrm{A}(-1,4,3)$ on the plane $\mathrm{P}: 2 x+\mathrm{m} y+\mathrm{n} z=4$, is $\left(-2, \frac{7}{2}, \frac{3}{2}\right)$, then the distance of the point A from the plane P, measured parallel to a line with direction ratios $3,-1,-4$, is equal to :

<p>$\left( { - 2,{7 \over 2},{3 \over 2}} \right)$ satisfies the plane $P:2x + my + nz = 4$</p> <p>$- 4 + {{7m} \over 2} + {{3n} \over 2} = 4 \Rightarrow 7m + 3n = 16$ ...... (i)</p> <p>Line joining $A( - 1,\,4,\,3)$ and $\left( { - 2,{7 \over 2},{3 \over 2}} \right)$ is perpendicular to $P:2x + my + nz = 4$</p> <p>$${1 \over 2} = {{{1 \over 2}} \over m} = {{{3 \over 2}} \over n} \Rightarrow m = 1$$ & $n = 3$</p> <p>Plane $P:2x + y + 3z = 4$</p> <p>Distance of P from $A( - 1,\,4,\,3)$ parallel to the line</p> <p>${{x + 1} \over 3} = {{y - 4} \over { - 1}} = {{z - 3} \over { - 4}}:L$</p> <p>for point of intersection of P & L</p> <p>$2(3r - 1) + ( - r + 4) + 3( - 4r + 3) = 4 \Rightarrow r = 1$</p> <p>Point of intersection : $(2,\,3,\, - 1)$</p> <p>Required distance $= \sqrt {{3^2} + {1^2} + {4^2}} = \sqrt {26}$</p>