Determine the value of $a^2+b^2+c^2$ if $a+b+c=14$ and $ab+bc+ca=45$.

The correct answer is 106.

Step 1: Identify the given values
We are given:
$a+b+c=14$
$ab+bc+ca=45$

Step 2: Recall the algebraic identity
The algebraic identity for the square of a trinomial is:
${(a+b+c)}^2=a^2+b^2+c^2+2(ab+bc+ca)$

Step 3: Substitute the known values
Substitute $a+b+c=14$ and $ab+bc+ca=45$ into the formula:
${14}^2=a^2+b^2+c^2+2\left(45\right)$

Step 4: Simplify and calculate
Compute the square of 14 and double 45:
$196=a^2+b^2+c^2+90$
Subtract 90 from both sides of the equation:
$a^2+b^2+c^2=196-90$
$a^2+b^2+c^2=106$

Therefore, the value of
$a^2+b^2+c^2$
is 106.