If $(a-18{)}^2+(b-12{)}^2+(c-6{)}^2=0,$ then find the value of $\sqrt{a+b+c}$

The correct answer is ±6.

We are given the equation:

$(a-18{)}^2+(b-12{)}^2+(c-6{)}^2=0$

Step 1: Apply the key property of squares.

We know a fundamental property: the square of any real number is always greater than or equal to zero. That is, for any real number x:

$x^2\ge 0$

This means each of the three terms in our equation is individually non-negative:

$(a-18{)}^2\ge 0$
$(b-12{)}^2\ge 0$
$(c-6{)}^2\ge 0$

Step 2: Conclude that each term must individually be zero.

Since the sum of three non-negative quantities equals zero, the only possibility is that each term is exactly zero. (If even one term were positive, the total sum would exceed zero — a contradiction.)

Therefore:

$(a-18{)}^2=0\Rightarrow a=18$

$(b-12{)}^2=0\Rightarrow b=12$

$(c-6{)}^2=0\Rightarrow c=6$

Step 3: Calculate a + b + c.

$a+b+c=18+12+6=36$

Step 4: Find the required value.

$\sqrt{a+b+c}=\sqrt{36}=\pm 6$

The value of $\sqrt{a+b+c}$ is ±6.