${?}^2\times 21-\sqrt{30\times 35\times 63\times 54}=189$

What will come in the place of question (?) mark in following questions.

The correct answer is $3\sqrt{11}$.

Let the question mark ($?$) be represented as $x$. The given mathematical equation is:

$21x^2-\sqrt{30\times 35\times 63\times 54}=189$

Step 1: Simplify the square root term $\sqrt{30\times 35\times 63\times 54}$

First, express each factor under the square root in terms of its prime factors:
• $30=2\times 3\times 5$
• $35=5\times 7$
• $63=3^2\times 7$
• $54=2\times 3^3$

Combining these inside the square root gives:

$\sqrt{(2\times 3\times 5)\times (5\times 7)\times (3^2\times 7)\times (2\times 3^3)}$

Group the exponential powers together:

$=\sqrt{2^2\times 3^6\times 5^2\times 7^2}$

Taking the square root of each perfect square term:

$=2\times 3^3\times 5\times 7$
$=2\times 27\times 5\times 7$
$=10\times 189=1890$

Step 2: Substitute the simplified square root back into the equation

$21x^2-1890=189$

Add $1890$ to both sides:

$21x^2=189+1890$
$21x^2=2079$

Step 3: Solve for $x$

Divide both sides by $21$:

$x^2=\frac{2079}{21}$
$x^2=99$

Taking the square root on both sides:

$x=\sqrt{99}=\sqrt{9\times 11}=3\sqrt{11}$

Thus, the value coming in place of $?$ is $3\sqrt{11}$.