Step 1

Given, point (6,-2,5) and normal vector j+2k

Step 2

Concept:

When point \(\displaystyle{\left({x}_{{{0}}},{y}_{{{0}}},{z}_{{{0}}}\right)}\) and normal vector ai+bj+ck, then equation of plane is given by \(\displaystyle{a}{\left({x}-{x}_{{{0}}}\right)}+{b}{\left({y}-{y}_{{{0}}}\right)}+{c}{\left({z}-{z}_{{{0}}}\right)}={0}\)

Step 3

Here, \(\displaystyle{\left({x}_{{{0}}},{y}_{{{0}}},{z}_{{{0}}}\right)}={\left({6},-{2},{5}\right)}\) and a=0, b=1, c=2

Then, equation of plane is 0(x-6)+1(y-(-2))+2(z-5)=0

(y+2)+2z-10=0

Equation of plane y+2z-8=0

Given, point (6,-2,5) and normal vector j+2k

Step 2

Concept:

When point \(\displaystyle{\left({x}_{{{0}}},{y}_{{{0}}},{z}_{{{0}}}\right)}\) and normal vector ai+bj+ck, then equation of plane is given by \(\displaystyle{a}{\left({x}-{x}_{{{0}}}\right)}+{b}{\left({y}-{y}_{{{0}}}\right)}+{c}{\left({z}-{z}_{{{0}}}\right)}={0}\)

Step 3

Here, \(\displaystyle{\left({x}_{{{0}}},{y}_{{{0}}},{z}_{{{0}}}\right)}={\left({6},-{2},{5}\right)}\) and a=0, b=1, c=2

Then, equation of plane is 0(x-6)+1(y-(-2))+2(z-5)=0

(y+2)+2z-10=0

Equation of plane y+2z-8=0