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It is given in the question that $P, Q, R$ are collinear points.

Collinear means lying on the same line which means that $P, Q, R$ lies on the same straight line.

Let us suppose that the co-ordinates of $R$ is $(x,y)$. Now, we have $P(7,7)$, $Q(3,4)$, $R(x,y)$ and also $PR=10$.

We have to use distance formula in order to find $PQ,QR,PR$. So, according to distance formula if $A({{x}_{1}},{{y}_{1}})$ and $B({{x}_{2}},{{y}_{2}})$ then $AB=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}$.

Now, $P,Q,R$ are three collinear points then they must satisfy,

$\therefore PQ+QR=PR$

Using distance formula, we get

$\Rightarrow \sqrt{{{(7-3)}^{2}}+{{(7-4)}^{2}}}+\sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=10$

$\Rightarrow \sqrt{16+9}+\sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=10$

$\Rightarrow 5+\sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=10$

\[\Rightarrow \sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=5\]

Squaring both sides, we get

$\Rightarrow {{(3-x)}^{2}}+{{(4-y)}^{2}}=25$

$\Rightarrow {{x}^{2}}+{{y}^{2}}-6x-8y+25=25$

Cancelling 25 from both sides, we get

$\Rightarrow {{x}^{2}}+{{y}^{2}}=6x+8y...................(1)$

Also, it is given that $PR=10$, using this we get

$\therefore PR=10$

\[\Rightarrow \sqrt{{{(7-x)}^{2}}+{{(7-y)}^{2}}}=10\]

Squaring both sides, we get

\[\Rightarrow {{(7-x)}^{2}}+{{(7-y)}^{2}}=100\]

$\Rightarrow {{x}^{2}}+{{y}^{2}}-14x-14y+98=100$

Putting the value of ${{x}^{2}}+{{y}^{2}}$ from equation (1), we get

$\Rightarrow 6x+8y-14x-14y=100-98$

$\Rightarrow -8x-6y=2$

Now, expressing $y$ in terms of $x$, we get

$\Rightarrow y=\dfrac{-(2+8x)}{6}=\dfrac{-(1+4x)}{3}..................(2)$

Putting the value of $y$ from equation (2) in equation (1), we get

$\Rightarrow {{x}^{2}}+{{\left( \dfrac{-(1+4x)}{3} \right)}^{2}}=6x+\dfrac{-(1+4x)}{3}\times 8$

$\Rightarrow \dfrac{9{{x}^{2}}+1+8x+16{{x}^{2}}}{9}=\dfrac{18x-8-32x}{3}$

$\Rightarrow 25{{x}^{2}}+8x+1=3(-14x-8)$

$\Rightarrow 25{{x}^{2}}+8x+1+42x+24=0$

$\Rightarrow 25{{x}^{2}}+50x+25=0$

$\Rightarrow 25({{x}^{2}}+2x+1)=0$

$\Rightarrow {{(x+1)}^{2}}=0$

$\Rightarrow x+1=0\Rightarrow x=-1$

Now, putting the value of $x$ in equation (2), we get

$\Rightarrow y=\dfrac{-(1+4\times -1)}{3}$

$\Rightarrow y=\dfrac{3}{3}=1$

So, $x=-1$ and $y=1$.