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**Hint:**Hint! Let the foot of perpendicular be p(a, b, c). Then directions ratio of OP are (a − 0, b − 0, c − 0) = (a, b c) then, direction ratios of the normal to a, b, c is

a(x − a) + b(y − b) + c(z − c) = 0

**Complete step by step solution:**Given that,

the foot of the perpendicular from the origin to a plane is (a, b, c).

Now,

According to the question.

⇒ Let the foot of the perpendicular be P(a, b, c).

Then,

⇒ direction ratios of OP are (a − 0, b − 0, c − 0) = (a, b, c).

So,

⇒ the equation of plane passing through P(a, b, c)

then,

the direction ratios of the normal to which are a, b, c is

$\text{d}\cdot {{\text{r}}_{8}}(x-{{x}_{1}})+\text{d}\cdot \text{r}(\text{y}-{{\text{y}}_{1}})+\text{d}\cdot {{\text{r}}_{8}}(\text{z}-{{\text{z}}_{1}})=0$

where,

${{x}_{1}}$ = x-co-ordinates of point P.

${{\text{y}}_{1}}$ = y-co-ordinates of point P.

${{\text{z}}_{1}}$ = z-co-ordinates of point P.

Now,

⇒ Here, \[{{x}_{1}}=\text{a}\], \[{{\text{y}}_{1}}=\text{b}\], \[{{\text{z}}_{1}}=\text{c}\].

So,

⇒ a(x − a) + b(y − b) + c(z − c) = 0

⇒ After multiplying, we get;

⇒ $\text{a}x-{{\text{a}}^{2}}+\text{by}-{{\text{b}}^{2}}+\text{cz}-{{\text{c}}^{2}}=0$

⇒ ax + by + cz = ${{\text{a}}^{2}}+{{\text{b}}^{2}}+{{\text{c}}^{2}}$.

Hence, the correct option is (C).

**Note:**In this type of question we know about the direction ratios that is If a, b, c are three numbers proportional to the direction cosine l, m, n of a straight line, then a, b, c are called its direction ratios..then, First, find out the direction ratios of line and then find direction ratios of the normal to point (a, b, c)

= $\text{d}\cdot {{\text{r}}_{8}}(x-{{x}_{1}})+\text{d}\cdot {{\text{r}}_{8}}(\text{y}-{{\text{y}}_{1}})+\text{d}\cdot {{\text{r}}_{8}}(\text{z}-{{\text{z}}_{1}})=0$

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