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

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

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

Let the unknown value represented by the question mark (?) be x.
The given equation can be written as:

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

Step 1: Simplify the square root term.
Break down each number inside the square root into its prime factors:
• 30 = 2 × 3 × 5
• 35 = 5 × 7
• 63 = 7 × 9 = 7 × 3²
• 54 = 6 × 9 = 2 × 3 × 3² = 2 × 3³

Now, group the factors together under the radical:

$\sqrt{30\times 35\times 63\times 54}=\sqrt{(2\times 3\times 5)\times (5\times 7)\times (7\times 3^2)\times (2\times 3^3)}$

Combining powers of the same base gives:

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

Take the square root of each perfect square factor:

$=2\times 3^3\times 5\times 7$

$=2\times 27\times 5\times 7=1890$

Step 2: Substitute the simplified value back into the original 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:

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

$x^2=99$

Take the square root of both sides:

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

Thus, the value that replaces the question mark (?) is 3√11.