According to ques,
plane is: \[2x\text{ }\text{ }2y\text{ }+\text{ }4z\text{ }+\text{ }5\text{ }=\text{ }0\] and point: \[\left( 1,\text{ }3/2,\text{ }2 \right)\]
The direction ratios of the normal to the plane are: \[2,\text{ }-2,\text{ }4\] Hence,
Equation of the line passing through (1, 3/2, 2) and direction ratios are equal to the direction ratios of the normal to the plane i.e. 2, -2, 4 is:
Since any point in the plane is:
\[2\lambda \text{ }+\text{ }1,\text{ }-2\lambda \text{ }+\text{ }3/2,\text{ }4\lambda \text{ }+\text{ }2\]
Also, the point lies in the plane, then
\[2\left( 2\lambda \text{ }+\text{ }1 \right)\text{ }\text{ }2\left( -2\lambda \text{ }+\text{ }3/2 \right)\text{ }+\text{ }4\left( 4\lambda \text{ }+\text{ }2 \right)\text{ }+\text{ }5\text{ }=\text{ }0\]
Or,
\[4\lambda \text{ }+\text{ }2\text{ }+\text{ }4\lambda \text{ }\text{ }3\text{ }+\text{ }16\lambda \text{ }+\text{ }8\text{ }+\text{ }5\text{ }=\text{ }0\]
Or,
\[24\lambda \text{ }+\text{ }12\text{ }=\text{ }0\text{ }\lambda \text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\]
Hence, the coordinates of the point in the plane are:
\[2\left( -1/2 \right)\text{ }+\text{ }1,\text{ }-2\left( -1/2 \right)\text{ }+\text{ }3/2,\text{ }4\left( -1/2 \right)\text{ }+\text{ }2\text{ }=\text{ }0,\text{ }5/2,\text{ }0\]
Therefore, the foot of the perpendicular is (0, 5/2, 0) and the required length